Mathematics - 2017-02-05 (page 2 of 3)

1:00 PM

@DHMO Can you solve this : If $mn$ coins have been distributed into $m$ purses, $n$ into each find the chance that two specified coins will be found in the same purse

I tried like: Total number of ways in which $mn$ coins can be distributed in $m$ purses with $n$ coins in each is:
$$\dfrac{(mn)!}{(n!)^mm!}$$
Suppose we have distributed any $n$ coins (containing the specified $2$ coins) in any selected purse, then the probability is: $$\dfrac{\binom{mn}{n-2}\binom{m}{1}\dfrac{(mn-n)!}{(n!)^{m-1}(m-1)!}}{\dfrac{(mn)!}{(n!)^mm!}}$$

However, my answer doesn't seem to be the correct one. Can you identify the mistake ? @DHMO

anonymous

1:24 PM

@DHMO After simplifying it comes out to be $$\frac{m(n-1)}{mn-1}$$ but the actual answer should be $$\frac{n-1}{mn-1}$$.

DHMO

@anonymous that's easy, change the (m-1)! to m!

anonymous

@DHMO Oh, I initially neglected that even the purse which contains the 2 specific coins has n balls. Yeah, then it should be $m!$. Thanks!!

Phew

I was tearing my hairs...LOL XD

DHMO

@anonymous you can see from the answer that there has to be a simpler way

anonymous

@DHMO Tell me the simpler way!

DHMO

the answer seems like $\dfrac {m \binom n 2} {\binom {mn} 2}$

you might want to consider only those 2 coins

anonymous

1:28 PM

@DHMO Umm, could you explain your thought process

DHMO

@anonymous I worked from the answer...

anonymous

I have no idea how we can directly conclude $\dfrac {m \binom n 2} {\binom {mn} 2}$. Will think...:P

It looks like selecting 2 coins from mn

at the denom

and selecting 2 coins from n coins

doesn't make sense

DHMO

let's ignore it

I might be talking nonsense

anonymous

yep..haha XD

I've some more probability questions

do you want to solve ?

DHMO

yes

anonymous

1:32 PM

okay..wait a min :D

@DHMO I solved this one but it was a good problem

You can try

(Probably it will be easy for you) :P

DHMO

12. (b)(2b)(3b) = 0.7 ((6b-3)C(3))
6b^3 = 0.7 (2b-1)(3b-1)(6b-1)
can i not solve the equation

ugh, i'm quite busy, sorry

anonymous

okay...no probs

we will continue later :)

DHMO

i thought you're asking me to help you, so

anonymous

@DHMO No no, I had solved it. I thought you were free and looking for good problems and that's why I gave it to you. If you are busy then no problem. See you later!

DHMO

sure, thanks

Astyx

2:03 PM

hi chat

Alessandro Codenotti

Hi @Astyx

Astyx

How are you ? @Alessandro

Andrzej Golonka

Hello. I'm looking for some keywords about what to read. How do I calculate area under a curve, for given initial and terminal time, in polar coordinates?

Alessandro Codenotti

I'm doing an unhealthy amount of physics but good apart from that @Astyx (I hope there are no physicists around :P)

How about you?

shmiggens

Hi chat.

Astyx

2:13 PM

@Alessandro Enjoying my first day of my "holidays" (which will consist mainly of revisions, but hey) :)

Hi @shmiggens

shmiggens

Hello @Astyx. Is there anyone here who understands probability theory?

Astyx

Ask your question, if someone can and wants to answer, she/he will

Alessandro Codenotti

My holidays are about to end (where holidays means lectures free but exams full time period)

I'm actually looking forward to the beginning of the semester, I'm going to have a few interesting courses

shmiggens

Here is the question math.stackexchange.com/questions/2129683/…

Astyx

Your semester starts late does it not ? Mine has begun one month ago

shmiggens

2:17 PM

As a first step, I think that it is clear that I should apply Kronecker's lemma, but this problem is difficult because the random variables are not necessarily independent.

Alessandro Codenotti

In the middle of February and I'll have lectures until the beginning of June more or less

Astyx

I don't get why the question was downvoted

Alessandro Codenotti

There are 14 weeks of lectures in a semester if I remember correctly, followed by exams

shmiggens

@Astyx I don't know why either.

Astyx

Well it's true my lectures end in april, because then the competitive exams start

Henning Makholm

2:33 PM

Getting rather irritated with authors of logic textbooks who insist on phrasing questions about propositional entailment in terms of "is this a valid argument?". NO, AUTHOR, a list of complex premises immediately followed by a conclusion is not a valid argument, even if the conclusion happens to be entailed by the premises -- because you did not include any argument for that entailment at all.

user21820

@HenningMakholm While I agree in principle, some authors take "argument" in the sense of Aristotle, and whose "validity" refers to logical entailment. As long as they spelt out clearly their meaning of "argument", they can't really be faulted in my opinion.

Semiclassical

Hi chat

Astyx

Hi @Semi @Balarka

Balarka Sen

Hi @Astyx @SemiC @Alessandro

Alessandro Codenotti

Hi @Balarka @Semi

Balarka Sen

2:41 PM

Still doing physics? :P

user21820

@BalarkaSen Can you really ping so many people in chat at once?

Balarka Sen

You can

Alessandro Codenotti

Yep @Balarka speaking of which do you have a moment for a question about Hamiltonian mechanics? @semi

Semiclassical

You can ask, no guarantees I can answer.

anonymous

@DHMO I found an easier way to solve the coin purse problem.One of the coins will certainly be in one of the $m$ purses; but who cares which.   Wheresoever that coin is, then there are $(nm{-}1)$ 'places' the other coin could be, of which $(n{-}1)$ are in the same purse as the first.   There is no need to involve factorials

Alessandro Codenotti

2:54 PM

It's a bit of a vague question: I have a physical system described by an hamiltonian $H$ and I want to find a function $F$ inducing a canonical transformation such that $H'=H+\frac{\partial F}{\partial t}$ is always $0$, when does such an $F$ exist? The professor always just said "let's take $F$ such that" without mentioning anything about its exsistence

Semiclassical

Hrm. I should know the answer to this.

DHMO

@anonymous nice

Semiclassical

This is Hamilton-Jacobi stuff?

Alessandro Codenotti

Yeah, we started with a "classical Hamiltonian" and we want to arrive at the Hamilton-Jacobi equations, so we talked about canonical transformations

Semiclassical

I wouldn't be surprised if we were similarly handwavey about whether such an F actually exists.

Alessandro Codenotti

3:06 PM

And concluded that after a canonical transformation the new hamiltonian looks like $H+\frac{\partial S}{\partial t}$ for some function $S$ (where $H$ is the old hamiltonian)

Semiclassical

Frankly, I don't know.

Hamilton-Jacobi theory was something I remember doing and not understanding what was really hahppening

I mean, something along the lines of "existence of solutions to HJ equation"

Alessandro Codenotti

I can't claim I understand what's going on either... To be fair the second half of the course will be about pdes so we might return to those solutions then

John Doe

Hey just checking if I have an inequality in which some terms are under an absolute value I can still divide everything by something that's strictly positive?

Astyx

3:22 PM

Yes

John Doe

Thanks

Luigi M

Hi all! I've a quick question about characteristic classes. does the relation X(W)=<e(W),[W,dW]> hold for oriented manifold W with boundary dW ?

where X(W) is the Euler characteristic

Balarka Sen

3:37 PM

@LuigiM That's an interesting question. Is e(W) defined for manifolds with boundary just like manifolds without boundary? Aka, cup square of the zero section of the projectivized tangent bundle of W. I think it should be true; if there is a vector field on W pointing outwards along $\partial$W, the sum of it's indices is X(W) - this is a version of Poincare-Hopf with boundary. That's kind of what you'd want here, I guess.

Hmm, e(W) should be cup square of the zero section of TW relative to the tangent space along the boundary or something like that. I don't know how to carefully handle stuff happening at the boundary.

Luigi M

@BalarkaSen to be honest, I need the weaker <w_5(W),[W,dW]> mod2. so the question would be, how are char. classes are defined on W,dW

@BalarkaSen I'm reading a paper where there is evaluation, and I would like to evaluate it in order to improve the result

Balarka Sen

I guess you could define them axiomatically but there should be some analogue of interpreting the Euler class as self intersection number of the zero section in a careful way.

I think e(W) is [W, dW] \cup [W, dW] in H^*(TW, N(dW)) where N(dW) is normal bundle to the boundary inside the tangent bundle along dW (which is isom to a trivial line bundle).

In which case you could apply the Poincare-Hopf with boundary to prove your claim.

@LuigiM It's best to consult someone more knowledgeable (@MikeMiller, perhaps?) about that though.

Luigi M

3:53 PM

@BalarkaSen that would be great, I will try thinking about it in the meanwhile. so maybe I will be able to post a precise question

Balarka Sen

Good idea.

becko

4:43 PM

hi all

Sha Vuklia

hi

is it true that if a function $f$ can be expressed as a series, that it's also automatically expressible as a Taylor series?

JessyunBourne

1

Q: Find a finite generating set for $Gl(n,\mathbb{Z})$

I need to find a finite generating set for $Gl(n,\mathbb{Z})$. I heard somewhere once that this group is generated by the elementary matrices - of course, if I'm going to prove that $GL(n,\mathbb{Z})$ has a finite generating set, I would need to prove that any matrix $M\in GL(n,\mathbb{Z})$ can ...

Nothing anyone's said so far is helpful. The answer is confusing and the person who posted comments is just snarky.

@arctictern could you help? :(

Liad

5:07 PM

@ShaVuklia i think no, because if $f$ can be expressed as a series, it does not mean f is differentiable (at all) .

Sha Vuklia

@Liad Oh sorry, I didn't say it correctly: I meant if $f$ can be expressed as a power series

Velvet

@ShaVuklia All analytic functions are smooth.

Sha Vuklia

@Velvet Smooth function implies expressible as Taylor series?

Velvet

@ShaVuklia No

Balarka Sen

Any power series can be written as a Taylor series. This is a theorem of Borel.

s.harp

5:13 PM

You have to be careful how to phrase that without being tautological

Velvet

For example the function $\begin{cases}f(x) = 0, \; x \le 0 \\ f(x) = \exp(x^{-2}), \; x > 1 \end{cases}$ is smooth but not analytic.

Balarka Sen

@Velvet But we're starting with a function which admits a power series in the first place, which your thing doesn't.

s.harp

The result is that inside the radius of convergence of a power series the function given by the power series has a taylor expansion at every point that converges to the function in some neighbourhood of that point

Velvet

@BalarkaSen I was replying to "Smooth function implies expressible as Taylor series".

Balarka Sen

Yeah, so what I meant is that your "all analytic functions are smooth" does not answer the original question.

Sha Vuklia

5:15 PM

Oh okay, so we usually aren't interested in the Taylor series, if we already know the series of the function. If I'm correct, Taylor series is a useful way for us to find the series of the function?

(Because it "is" the series)

s.harp

(There was almost a fight in the library 10 minutes ago, people threatening each other with going to the police etc, so weird)

Sha Vuklia

wow what kind of library do you study atXD

I've never experienced violence in a library:P

s.harp

they weren't getting violent, just progressively louder

Sha Vuklia

yea ok that's annoying:d

Astyx

What's the difference between a power series and a Taylor series ?

And hi to those I have not seen today

s.harp

5:24 PM

@Astyx my guess is in the way it is used, a Taylor series is a power series associated to a smooth function

Astyx

But in the end it's pretty much the same right ?

Balarka Sen

@s.harp I wonder if it's easier than Borel's theorem though. There should be a result about termwise differentiation somewhere; once one can do that you'd be done.

Uniform convergence of the partial sums is not sufficient.

Astyx

It is on any compact

And since differentiation is a local property, it's sufficient

Balarka Sen

What? Smooth functions $f_n$ converging to $f$ uniformly does not mean $f_n'$ converges to $f'$, or even that $f$ is differentiable.

s.harp

@BalarkaSen I think I misunderstood your statement then, I thought you were saying every power series with non-zero radius of convergence is its own Taylor series / is the Taylor series of a smooth function

Astyx

5:29 PM

^

Velvet

Is there a term for a metric space $(X, d)$ such that for any $x \in X$ there exists $y \in X, \; x \neq y$ such that $d(x, y) < \epsilon, \; \forall \epsilon \in \mathbb R^*_+$?

Balarka Sen

@s.harp That's of course not what Borel says; it just says coefficients of every power series comes from Taylor coefficients of some smooth function.

Ramanujan

7 hours ago, by Ramanujan

$$\left ( \begin{matrix}
3 & 10 & 5 \\
4 & 6 & 2 \\
\end{matrix} \right )
\left ( \begin{matrix}
a\\
b\\
c\\
\end{matrix} \right ) =
\left ( \begin{matrix}
0 \\
0\\
\end{matrix} \right )
$$

7 hours ago, by Ramanujan

What will be $\Delta_1$ for this matrix? (In Cramer's rule)

Any idea?

Astyx

What's $\Delta_1$ ?

Velvet

^

Ramanujan

5:38 PM

In camera rule a= delta_1/delta

Astyx

It can't apply here

Ramanujan

I asked this question today…

0

Q: How to get ratio of a,b,c from 2 equations in a,b,c

I have 2 equation in terms of $a,b$ and $c$ . $3a + 10b + 5c =0$ and $4a + 6b + 2c =0$ I need to find a:b:c and answer is $\dfrac a5 = \dfrac {b}{-7} = \dfrac {c}{11}$ I want to know how to get that? My attempt: Given equations can be written in form $$\left ( \begin{matrix} 3 & 10

matrices systems-of-equations ratio

Astyx

If there is a solution, there can't only be one

Ramanujan

I am at end of it but but unable to find Delta _1

Astyx

How far are you in Linear Algebra ?

Have you studied abstract vector spaces yet ?

Velvet

5:40 PM

Isn't Cramer's rule only applicable to square matrices?

Astyx

Yes it is

Invertible ones even, IIRC

Velvet

IIRC?

Astyx

If I recall correctly

Velvet

Ah, I'm not good with abbreviations. xD

According to my book it has to be invertible.

Ramanujan

@Astyx no,I didn't learnt eigen vectors (if I pronounced correctly)

Astyx

5:44 PM

eigenvectors ?

Ramanujan

Yes,that

Velvet

Ugh, Cramer's rule is so ugly.

Astyx

Is it ?

Velvet

Well at least the way how they introduce it in this book doesn't really cry for elegance.

Astyx

Is it not simply the determinant of the matrix where the k-th collumn is replaced by the result over the determinant of the matrix

Ramanujan

5:46 PM

To many students anything that isn't a small whole number is ugly.

XD

Astyx

@Ramanujan You don't need eigenvectors to do that

Velvet

@Astyx yes

Ramanujan

OK,so let's proceed

Astyx

But knowing about abstract vector spaces, linear applications ... helps

Ramanujan

So…?

Velvet

5:49 PM

@Ramanujan What's the problem?

Astyx

Well wrtie the system of equations linked to the matrix equality

Ramanujan

0

Q: How to get ratio of a,b,c from 2 equations in a,b,c

I have 2 equation in terms of $a,b$ and $c$ . $3a + 10b + 5c =0$ and $4a + 6b + 2c =0$ I need to find a:b:c and answer is $\dfrac a5 = \dfrac {b}{-7} = \dfrac {c}{11}$ I want to know how to get that? My attempt: Given equations can be written in form $$\left ( \begin{matrix} 3 & 10

matrices systems-of-equations ratio

Astyx

Oh you already have

Velvet

Form a matrix, use Gaussian elimination, done.

Ramanujan

@Velvet I want a:b:c not a=? b=? And c=?

Astyx

5:51 PM

second equation can be divided by 2

Velvet

@Ramanujan What's a:b:c?

Astyx

Then you can get a in terms of b and c by substracting

Ramanujan

Thanks,but it is given that a:b:c is $$ \dfrac {a}{30-20} = \dfrac{b}{6-20} = \dfrac {c}{-18 + 14}$$ which is like denominators are determinant of cofactor matrices and I think it has something to do with matrices. Can you show how only using matrix? — Ramanujan 8 hours ago

Astyx

And finally remove a in one of the equations

Velvet

That only reminds me of homogeneous coordinates in projective spaces.

Astyx

5:53 PM

Oh sorry

You can check that (77, -55, 35) works

And argue that the ker is of dimension 1 by rank theorem

Ramanujan

Rank is 2

Astyx

So you know about rank

How ?

Ramanujan

Det of 2×2 matrix is not equal to 0

Astyx

Huh

Right

Velvet

So for all $a, b, c, d \in \mathbb C$ $ad-bc \neq 0$? xD

Ramanujan

5:57 PM

@Astyx it dont work because it is a/(+term)=b/(- term) = c/(- terms)

Astyx

Actually (77, -55, 35) doesn't work

Check $(5, -7, 11)$

Ramanujan

@Astyx that's the only simplified ratio :)

Astyx

Yes

And once you know it works, since the dimension of the ker is 1, any root is in the span of this vector

Which allows you to conclude

Ramanujan

Let me have clear view of questions , in my text book when equation are given like ax+by+cz=0 and dx + ey+fz = 0 then ratio is $$\dfrac {x}{bf - ec} = \dfrac{y}{at - dc} = \dfrac{z}{ae - bd}$$

So in general how they get all denominators?

@Astyx ^

Astyx

I guess you can check it works

And if the formula works, then the rank is 2 so the ker is 1, QED

Semiclassical

6:11 PM

I'd think the simplest way is to rewrite the equations to $ax+by=-cz$ and similarly for the other equation.

Ramanujan

Ker?

Semiclassical

Which is a full rank system of equations in x,y.

Astyx

Semi's right

Velvet

@Ramanujan Kernel

Ramanujan

@Semiclassical yes,but I am want to know how directly it got relation

Velvet

6:12 PM

I.e. all vectors $x$ such that $f(x) = 0$.

Semiclassical

Use Cramer's rule on that x,y system. You'll get solutions for $x,y$ which you can rearrange to get the stated relation.

Ramanujan

Wooh, I have proceeded somewhat. Now from Cramer's rule $a= \dfrac {\Delta_1}{\Delta} , b = \dfrac {\Delta_2}{\Delta} , c= \dfrac {\Delta_3}{\Delta} $ which gives $$\dfrac {a}{\Delta_1} = \dfrac {b}{\Delta_2} =\dfrac {c}{\Delta_3} = \Delta $$

Now my only problem is what would be $\Delta_1$ for a 2×3 matrix.

@Semiclassical ^

Semiclassical

It's not for a 2x3 matrix. I'm saying to use Cramer's rule on the 2-by-2 system formed by moving $z$ to the other side and treating it as the inhomogeneous part.

Ramanujan

original question

Semiclassical

(This approach has one flaw, namely that it forces one to break the symmetry between x,y,z. Not sure I see a good way to avoid that.)

My approach is the same as that of lab bhattacharjee and David Quinn in those answers, btw.

I like Hagen von Eitzen's approach best, though, which is to interpret the problem geometrically (get set of vectors which are orthogonal to two given vectors by using the cross-product).

Ramanujan

6:19 PM

Can you help me with my approach? (So that I can always use it :) )

it will save some time :P

Velvet

@Ramanujan You can't have a $\Delta_i$ for a non-square matrix as stated earlier.

Ramanujan

So can we introduce Row_3 as 1&1&1 ?

Semiclassical

If you want to get a specific $(x,y,z)$, sure.

Just recognize that you're adding an extra constraint to the system, one which you need to know how to remove at the end.

Ramanujan

Remove by removing one of solutions x=y=z=0 ?

Semiclassical

Actually, wait. I may be talking nonsense.

Yeah, that doesn't actually make sense as I say it.

Introducing a 111 row would amount to requiring x+y+z=0 as well. And that limits your solution set to just the trivial (0,0,0) solution.

Ramanujan

6:28 PM

Then we will get only one solution x=y=z=0 ?

Velvet

Well, back to studying.

Semiclassical

Quite. Which means that's the wrong thing to do. (Your original system definitely doesn't just have the trivial solution.)

Shobhit

hello guys

need help with a question : math.stackexchange.com/questions/2130202/find-null-space-of-t3

Semiclassical

@Shobhit Can't $(a,b,c)\to T(a,b,c)$ be written as a matrix multiplication?

Kirill

6:49 PM

Hello gentlemen! I have the linear function
$ \phi: \mathbb{R}^2 \to \mathbb{R}^2$ with $ \phi(
\begin{pmatrix}
1 \\ 3
\end{pmatrix}
) =
\begin{pmatrix}
-2 \\ -1
\end{pmatrix}
$
and
$ \phi(
\begin{pmatrix}
2 \\ 1
\end{pmatrix}
) =
\begin{pmatrix}
6 \\ 3
\end{pmatrix}
$
and have to give
$ \phi(
\begin{pmatrix}
x \\ y
\end{pmatrix}
)
$
for all
$
\begin{pmatrix}
x \\ y
\end{pmatrix}
\in \mathbb{R}^2$
using some kind of linear continuation. Without giving me a solution, could you please give me an idea about how it should be done? The book says I can define a function using the function values …

Fargle

7:50 PM

@Kirill Try finding $\phi(1,0)$ and $\phi(0,1)$. Then $\phi(x,y) = x\phi(1,0) + y\phi(0,1)$.

Balarka Sen

@Fargle What're you working on

Kirill

@Fargle I have
$ \phi(\begin{pmatrix} x \\ y\end{pmatrix} ) = a_1 \cdot \begin{pmatrix} -2 \\ -1 \end{pmatrix} + a_2 \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix}$ for $\begin{pmatrix} x \\ y \end{pmatrix} = \sum_{i=1}^{2}a_iw_i$, but am not sure that the solution should look like this.

Fargle

@BalarkaSen nothing

Kirill

@Fargle will work through your advice, thank you

Balarka Sen

@Fargle what are you supposed to work on

/ what were you working on

Fargle

8:02 PM

@BalarkaSen nothing

Royden exercises, most recently

Balarka Sen

ah, analysis.

maybe I should really ask where you are in analysis

Sha Vuklia

8:19 PM

hey guys

"intuitive" question here

how do we know the range of a Taylor expansion?

say we're considering $\sin(x)$ for $x\in \mathbb R$

arctic tern

you mean the range of the function defined by the taylor expansion? I see no reason it could be determined just by looking at the coefficients. taylor expansion is local, range is global.

Sha Vuklia

we know that $\sin(x)=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}x^{(2n+1)}$

how do we know here for what interval the maclaurin series equals $\sin x$?

arctic tern

ratio test indicates radius of convergence is infinity

Sha Vuklia

but that only means that the Taylor series converges

it doesn't have to mean that it equals $\sin x$

right?

Semiclassical

You'd need to have some other definition of $\sin x$ in that case.

arctic tern

8:23 PM

it will equal sin(x) on the whole of R

Semiclassical

Either you define $\sin x$ via that convergent Taylor series, or you pick some other definition.

Sha Vuklia

okay maybe another example

assume we have a smooth function that is not analytical

we can make a taylor series about one point

like the following example

arctic tern

sure, but it won't converge to it on an interval around the point if it's not analytic at that point

Sha Vuklia

yea exactly

so two questions:

arctic tern

sin(x) is analytic everywhere

Sha Vuklia

8:25 PM

1) do we have to check by plugging in values that the taylor series doesn't converge on the interval, or can we prove it sometimes

Semiclassical

Except at infinity, but that's exactly what we're not looking at.

Sha Vuklia

2) if it does converge on an interval around the point, does that mean that the radius of convergence equals the interval of the taylor series where it equals the function? so in other words, as soon as the taylor series equals a function on an open interval, does that mean the taylor series is valid throughout its entire radius of convergence?

Kirill

@Fargle I have got $\phi(1,0) = (4,2)$ and $\phi(0,1) = (-2,-1)$ and it seems to work

@Fargle but should the function not to be injective? I have $\phi((0,1)) = \phi((1,3))$.

@Fargle btw, what was the main idea? I should take the standard basis of the $\mathbb{R}^2$, find its images and this is it? I have got the given vectors $(1,3)$ and $(2,1)$ and have checked out that they are also a basis. Could I use this one to get the solution? I do not really felt it through, why should I do like this :)

Fargle

@Kirill Note that your image vectors are linearly dependent; therefore the function should not be injective.

Kirill

@Fargle I have noted that. Could you please explain it more? The image vectors are linear dependent, so the function is not injective. Why? The implication?

@Fargle btw thank you for helping me

Fargle

8:41 PM

@Kirill You're mapping two linearly independent vectors onto the same line--therefore, every vector has to go onto that line.

Sha Vuklia

i think i have an answer to my question now

Kirill

@Fargle I haven't worked with vectors graphically yet, only theoretically

Sha Vuklia

one theorem i know says the following; if all the derivatives are bounded by a single constant on some interval $(a,b)$, then we know that the remainder goes to 0 as $n$ goes to infinity, and then we know that the function equals its Taylor series on that interval

so for the maclauring series of $\sin x$, $\cos x$, $e^x$, and some combinations, we can apply this theorem to show the interval

i wonder if it it is possible for a function to be completely analytical over $\mathbb R$, yet it's taylor expansions all have a finite radius of converge. by the definition of analytical, it should be no problem if this is the case, because we only need an open interval about the point in question. i'll ask in the forum i guess

Kirill

@ShaVuklia it is hard to understand what you mean, if we are in $\mathbb{C}$

:)

Sha Vuklia

can we restrict ourselves to the real case then?

because i haven't worked with $\mathbb C$ much

or is my question not clear?

Balarka Sen

8:49 PM

does 1/(1-x) work

Sha Vuklia

i'll try it out, but i'm very slow:d