Mathematics - 2024-12-06

Claudio

00:07

so since the $g_k$ are nonzero in disjoint subsets $E_k$, the sup should be attained when $x \in E_{k_{min}}$, namely $E_{n+1}$ and so $$ \sup_{x \in [0,1]}\left\vert \sum_{k = n+1}^{n+p} g_k(x) \right\vert = \frac{1}{(n+1)+1} = 1/(n+2)$$

and this goes to zero so we have uniform conv

leslie townes

that feels like the right answer

note that for my more general problem (assuming disjoint subsets E_k) you will have pointwise convergence no matter what the sequence c_k is, and uniform convergence when the sequence goes to 0

Claudio

@SineoftheTime this key observation was enlightning hahaha, I didn't notice this

psie

If a real-valued function $F:(a,b)\to\mathbb R$ is absolutely continuous on every compact subinterval of $(a,b)$, is then the set, on which the derivative of $F$ does not exist, a null set? If $F$ is absolutely continuous on $[a,b]$, this is the fundamental theorem of calculus. But here we only have absolute continuity on every compact subinterval.

Claudio

uniform convergence is indeed a real pain in the ass :p

@leslietownes wait I can't see it yet

leslie townes

psie: is there any relation you can think of between the set on which the derivative of F does not exist and the subsets of [a + 1/n, b - 1/n] on which the derivative of F does not exist (for n large enough so that this makes sense)

Sine of the Time

00:14

@Claudio uniform convergence is the fun part :)

Claudio

@leslietownes maybe if the sequence isn't decreasing to zero then you can't estimate the sup easily, if you take $c_k = k$ then in each interval the sup increases

no ok you're right hahaha

leslie townes

yes sup |sum_{k=N}^infty g_k(x)| is just sup_{k >= N} |c_k| whatever that is

it may not be "easy" to compute

Claudio

@SineoftheTime I see you like to suffer

leslie townes

but if its not going to zero you have a problem

Claudio

@leslietownes yeah I made a stupid observation

I can finally write down the solution :) thanks @leslietownes @SineoftheTime you were of great help

Sine of the Time

00:20

@Claudio leslie's car does not have the left turn signal because he's always right

Claudio

No caps detected

gotta admit that's a nice one @SineoftheTime I might have to store this in the bag for the future ahahah

Sine of the Time

pretty sure I've read it somewhere ahah

psie

00:44

@leslietownes ok, thanks. I argued by contradiction. If the set on which F' does not exist is not a null set, then it has positive measure 0<e<(b-a). But then we can find n large enough so that 0<2/n<e<(b-a) and note, 2/n is the sum of the measures of (a,a+1/n) and (b-1/n,b). So F' might not exist on these sets plus on some null set in [a + 1/n, b - 1/n]. So in total the set does not have measure e, but <2/n. Contradiction. So e=0.

Thorgott

01:08

@BenSteffan maybe the proposition i claimed earlier is slightly incorrect. the thing i can demonstrate is that the subcategory $BiCart(\Delta^1)$ of $CoCart(\Delta^1)$ is closed under limits. under straightening-unstraightening, this gives a certain subcategory of $\mathrm{Fun}(\Delta^1,\mathbf{Cat}_{\infty})$ closed under limits whose objects are precisely the left adjoints, but it's not so clear to me what the morphisms of this essential image are...

copper.hat

@psie you are taking a difficult route. the union of a countable number of null sets is still null. take the union of the null sets in leslie's comment.

Thorgott

the morphisms in the essential image do satisfy the condition that their components are also compatible with the corresponding right adjoints up to natural equivalence, but that condition might be strictly weaker

the potential issue being that the right adjoints themselves don't necessarily "see" the choices of cartesian lifts that were used to define them

psie

@copper.hat ok, so you mean, since $\bigcup_1^\infty [a + 1/n, b - 1/n]=(a,b)$, the null sets will also be countable?

copper.hat

no. on each of those sets there is a null subset $E_n$ where $F$ is not differentiable. then $\cup_n E_n$ is also null and $F$ is differentiable everywhere else.

on $(a,b)$.

psie

ah ok, yes, that's much better

thank you :)

copper.hat

01:13

the null sets are not necessarily countable

but they have measure zero, which is the point

psie

👍

Ben Steffan

@Thorgott I'm intrigued. How do you get that the objects of the subcategory are left adjoints?

Ben Steffan

01:32

Is it just "you take the functors in the other direction obtained from Cartesian straightening and then unravel the definitions to show that these are adjoint?"

Nasser

07:46

Math experts: At school I learned that inverse Laplace transform of exp(-a s) is dirac delta(a-t)  for positive a. but now I asked google and the google AI tells me that inverse Laplace transform of exp(-a s)  is exp(a t) ! Below is screen shot.   To see this yourself, if your google search gives AI answer, just type

           what is inverse laplace of exponential?

In the search bar.

Math Guy

Hey everyone, I wanted to ask why, most of the time, do I not get much votes on answers I think are well-written and provide a quick/nice solution? I have seen similar answers by other high rep users that have garnered tens of upvotes, even if it’s just a simple trick!

Is this an indication of something wrong, or is it perfectly fine?

Nasser

Mathematica gives DiracDelta[t-a] but Maple gives Dirac delta (a-t). But the point is the inverse laplace transform of exp(-as) should not be exp(at). Google seem to have a bug in its AI !

nickbros123

@MathGuy i think its more about how popular the question gets, rather than anything answer specific

Jakobian

08:07

@Nasser sounds like you misunderstand how AI works

Nasser

@Jakobian I am sure I did. I was told that AI is now very powerful and can solve all sort of math problems !

I want to use it to help solve HW's

Jakobian

Couldn't be further from truth. AI sucks at math

Its a language model - what's its good at is producing seemingly valid and coherent sentences about a given subject

It might help you solve homework, but I wouldn't expect anything from it

Jakobian

08:28

reddit.com/r/learnmachinelearning/comments/18ck15r/…

psie

11:01

Given some disjoint, finite number of open intervals $(a_1,b_1),\ldots,(a_n,b_n)$. Under what condition(s) on $T:\mathbb R\to\mathbb R$ is it true that also $(T(a_1),T(b_1)),\ldots,(T(a_n),T(b_n))$ are disjoint?

There are two things I'm a bit wary about. First, is $T((a_k,b_k))=(T(a_k),T(b_k))$? I feel like it needs to be monotone for this. Second, are two pairs of intervals disjoint under $T$? Only if it preserves intersections, right? So some kind of injectivity, I think.

Jakobian

@psie that $T$ is strictly monotone

psie

nice, ok 👍 then it will be injective, so disjointness comes for free :)

Jakobian

not exactly

the intervals $(T(a), T(b))$ are not images of $(a, b)$ (understood as $\emptyset$ when $T(a) = T(b)$ and $(T(b), T(a))$ when $T(b) < T(a)$, I suppose)

although I suppose if you understand them this way then you can just demand that $T$ is monotone, without injectivity

Now lack of monotonicity should be equivalent to existence of $a < b < c$ such that $T(a) < T(b) > T(c)$ or $T(a) > T(b) < T(c)$. Then $(T(a), T(b))$ and $(T(b), T(c))$ won't be disjoint

so if we understand those intervals in this more general way, then this is equivalent to $T$ being monotone

psie

11:21

@Jakobian ok. I can give you some more context behind it. I'm trying to show if $F$ is absolutely continuous on every compact subinterval on $(a,b)$ and $F'$ is increasing (where it is defined), then $F$ is convex.

Let $x, y \in(a, b), \lambda \in(0,1)$ and suppose that $x<y$. Moreover, let $T:[x, y] \to[x, z]$ be given by $T(t)=\lambda t+(1-\lambda) x$.

Here is where I'd like to say that $F \circ T$ is absolutely continuous, because if $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right), \ldots,\left(a_{n}, b_{n}\right) \subset[x, y]$ is a finite collection of disjoint intervals, then so should $\left(T\left(a_{1}\right), T\left(b_{1}\right)\right),\left(T\left(a_{2}\right), T\left(b_{2}\right)\right), \ldots,\left(T\left(a_{n}\right), T\left(b_{n}\right)\right) \subset[x, z]$ be.

And $$\sum_{1}^{n}\left(T\left(b_{k}\right)-T\left(a_{k}\right)\right)=\sum_{1}^{n}\left(\lambda b_{k}-\lambda a_{k}\right)=\lambda \sum_{1}^{n}\left(b_{k}-a_{k}\right)<\sum_{1}^{n}\left(b_{k}-a_{k}\right).$$

Thorgott

@BenSteffan yeah, to be honest where I learned it (Land's book) "the cocartesian fibration classifying this functor is bicartesian" is used as definition of a functor admitting a right adjoint

psie

By the way, $z=\lambda y+(1-\lambda)x$.

Thorgott

if you have $y$ in the fiber over $1$, you can first pick a cocartesian lift $Ly\rightarrow y$ of $0\rightarrow 1$ and then a cartesian lift $Ly\rightarrow RLy$ of $0\rightarrow1$ and that gives you a $2$-horn that can be completed by cartesianness to yield you the component of the unit transformation $y\rightarrow RLy$ in the fiber over $1$

and, really, you can carry this out parametrically to obtain the bona fide unit transformation (or dually the counit transformation)

Jakobian

@psie If you interpret this how I wrote it, you can still make this work for more general $T$ as long as $T$ is Lipschitz

that is, if $F$ is absolutely continuous and $T$ is Lipschitz then $F\circ T$ should be absolutely continuous

If $L$ is the Lipschitz constant of $T$, and if sum of lengths of $(a_k, b_k)$ is $\leq \delta$, then sum of lengths of $(T(a_k), T(b_k))$ is $\leq L\delta$, so if you demand that $\delta'$ is like in the definition of absolute continuity of $F$ for $\varepsilon$, then if sum of lengths of $(a_k, b_k)$ is $\leq \delta'/L$, then sum of lengths of $(F(T(a_k)), F(T(b_k)))$ will be $\leq \varepsilon$

psie

11:37

ok 👍

Jakobian

Lipschitz and monotone of course, so we can have that they are disjoint

Sine of the Time

13:23

@MathGuy this is normal

Xander Henderson

@MathGuy If you are answering questions for the clout, you are in the wrong place. I would recommend that you simply ignore the voting entirely.

Sine of the Time

@xander

think_meaning_buildß

time is money during an exam

Sine of the Time

13:40

@think_meaning_buildß Xander does not like expressions like obvious, clear, trivial :\

think_meaning_buildß

No mathematician does.

Math Guy

@XanderHenderson thanks for the advice, Mr. Xander Henderson. No, I am not writing answers merely for more upvotes, I do it because I love it. It gives me satisfaction. I felt that I should get feedback as per what I deserve, that's all.

ILikeMathematics

14:01

Does this look fine? The last sentence is worded a bit badly, I think

Apparently, there is a mistake in this though

Xander Henderson

@MathGuy Wrong honorific. If you are going to use honorifics, use the correct ones.

And noöne "deserves" upvotes. Or, frankly, any kind of feedback at all. If you post an answer and it gets upvotes, cool. If not, cool. If it gets downvotes, cool. Don't worry about it.

Thorgott

1. there appears a $b_n$ without that being ever defined
2. (most importantly) the conclusion in the penultimate sentence is incorrect. you cannot infer $a_{n+1}=a_n$, because the $N$ you have chosen depends on $\varepsilon$!

Sine of the Time

why $b_n$ if you only used it once?

write directly $|a_{n+1}/a_n-1-0|$

Xander Henderson

1. What is $b_n$?
2. $N$ depends on the choice of $\varepsilon$, so the conclusion that $a_n = a_{n+1}$ for sufficiently large $n$ seems not to follow.

Oops... too slow.

Thorgott

also, generally speaking, your argument only used that the terms tend to $0$, which is much weaker than the actual assumption that the series converges, so that should raise some eyebrows

trying some simple examples shows that $\frac{a_{n+1}}{a_n}-1$ can very well tend to $0$ even if $(a_n)_n$ is unbounded

Math Guy

14:11

@XanderHenderson I'm sorry if I offended you, could you please tell me the correct honorific to use?

@XanderHenderson ok, noted. Thank you for the advice once again.

Xander Henderson

@MathGuy "Dr". Though, honestly, again, I prefer not to stand on ceremony. "Xander" is fine. But if you are going to use honorifics, use the right one.

Math Guy

@XanderHenderson ok, I'll keep that in mind.

ILikeMathematics

14:23

Ah, thanks

ILikeMathematics

14:36

@XanderHenderson Yep, just noticed it. Do you see an alternative approach to this? It seems just the fact that the thing in the sum tends to 0 is a bit weak

Jakobian

15:38

@XanderHenderson prof. Dr. Xander "The Greatest of All Time" Henderson

Hmm... shouldn't it be Alexander if its official

nickbros123

Is the set of all cycles in S4 that are of the type (ab)(cd) identifiable with Klein 4 ?

Ok I think yes unless I'm making a grave calculation error

Soumik Mukherjee

Yes

Thorgott

what do you mean by "cycles of type (ab)(cd)"

your notation suggests a double transposition, not a cycle

Soumik Mukherjee

But you need to add the identity

nickbros123

16:22

@Thorgott yeah I meant a permutation

Thorgott

yeah, then it's as Soumik says

nickbros123

17:06

ah yes, of course, identity

psie

Consider part (a) of the above exercise. I'm a bit confused about how to show these measures equal. We are working with the Borel $\sigma$-algebra $\mathcal B([c,d])$. If it were $\mathbb R$ instead of $[c,d]$, I'd show these two measures agree on the algebra of finite disjoint union of h-intervals, which are of the form $(x,\infty)$, $(x,y]$ and $\varnothing$ for $-\infty\leq x<y<\infty$.

But now I don't know what the algebra is that generates $\mathcal B([c,d])$. If I could work with compact intervals $[x,y]$, the image of these intervals under $G^{-1}$ would also be compact by continuity. Appreciate any help.

Jakobian

17:37

@psie take those h-intervals you were talking about and intersect them with $[c, d]$

$\mathcal{B}(Y) = \mathcal{B}(X)\cap Y$ for $Y\subseteq X$ a subspace of $X$

$\sigma(\mathcal{A}\cap Y) = \sigma(\mathcal{A})\cap Y$

psie

@Jakobian ah nice, didn't know. That's useful. So I have to show the measures agree on $(x,\infty)\cap [c,d]=[c,x]$, $(x,y]\cap [c,d]=[c,x]$ or $(x,y]\cap [c,d]=(y,d]$ or $(x,y]\cap [c,d]=[c,x]\cup (y,d]$ and finally $\varnothing$, which is trivial. So in total, show the measures agree on compact subintervals of $[c,d]$ and half open intervals $(x,y]\subset[c,d]$ and I should be good.

Ben Steffan

17:53

@Thorgott I see

that's pretty cool :)

Thorgott

yeah, it was very eye-opening when I first learned it

Ben Steffan

18:11

I think I made a good decision acquiring a physical copy of Land's book

Jakobian

@psie I don't think you took some of those intersections right, but yeah that's the idea

psie

@Jakobian for $I$ a subinterval of $[c,d]$, how do we know $G^{-1}(I)$ is an interval? Continuity?

Definition of interval; for any $x<y$ in $G^{-1}(I)$, $z\in (x,y)$ is also in $G^{-1}(I)$.

I think it suffices to show the measures only agree on half open intervals $(x,y]$. If they do, then we can use the regularity definition of Lebesgue measure in terms of covers of half-open intervals to rewrite the half-open intervals in terms of $G^{-1}$ and hopefully arrive at something that resembles the regularity definition of $\mu_G$.

Jakobian

@psie A continuous increasing function $G:[a, b]\to \mathbb{R}$ is a homeomorphism between $[a, b]$ and $[G(a), G(b)]$

Thorgott

@BenSteffan well, it has a ton of typos still, but yeah

I really recommend the 5th chapter

Jakobian

If $I\subseteq [G(a), G(b)] = [c, d]$ then $G^{-1}:[c, d]\to [a, b]$ being a continuous function, it maps $I$ to an interval

psie

18:24

@Jakobian ok, didn't know that the inverse is then also continuous...hmm.

Jakobian

Yeah. You could treat this as "1-dimensional invariance of domain theorem" but although it is a particular variant of "invariance of domain", its easier to prove

This should be in Spivak

by that I mean one of his calculus books

think_meaning_buildß

19:05

@Jakobian Dr. X GOAT H

Xander Henderson

19:16

@Jakobian I'm not a professor. That isn't the title I have.

Officially, I am "faculty of mathematics".

Sine of the Time

you're not a professor, but you're still the goat

2

Ben Steffan

goatendieck

think_meaning_buildß

of this chat

Xander Henderson

To be fair, a lot of faculty here use the title "professor", and the term "professor" is far less protected in the US academic system than it is in most of Europe, but my advisor was French, and I am super uncomfortable being called "professor".

Ben Steffan

Prof.*

*-essional mathematician

Jakobian

19:22

@XanderHenderson I know, professor Xander

Sine of the Time

Prof (iterole)

Ben Steffan

yum

Prof(inite space)

Jakobian

@XanderHenderson we often call teachers "professors" here too, although I don't really enjoy that either

Sine of the Time

We call high school teachers professors

think_meaning_buildß

There's a user on Academia.SE who truly deserves the GOAT title in terms of their rep points accumulation.

Xander Henderson

19:26

@SineoftheTime Gross.

@think_meaning_buildß XP is dumb.

think_meaning_buildß

:(

74

Q: Thank you Dr. Buffy, Academia.SE's GOAT

Yesterday, Buffy officially became Academia.SE's user with the greatest all-time reputation score: nearly 169K reputation! In just 2.5 years, Buffy has authored a stunning 3,742 answers and earned 48 gold badges, among many other positive contributions. Academia.SE is lucky to have many highly ac...

discussion

Jakobian

don't we have someone with over 500k rep points on math.se

Sine of the Time

yeah

Jakobian

well, I suppose math.se is bigger

Sine of the Time

Brian Scott

think_meaning_buildß

19:28

He used to be a chatroom regular

Sine of the Time

And Andre Nicolas

last seen 8 years ago

ModularMindset

I have the least rep in this chat - so what does that say about me?

Jakobian

Brain Scott, one of the experts in topology

I wonder if he fixed that construction I was asking him about in one of his answers

ModularMindset

I've been earning real life points

think_meaning_buildß

life doesn't give points

Jakobian

19:31

5

A: $Y$ is $T_1$ iff there is regular space $X$ s.t. all continuous function from $X$ to $Y$ is constant

Eric van Douwen’s paper ‘A regular space on which every continuous real-valued function is constant’, Nieuw Arch. Wisk. $30$ $(1972)$, $143$-$145$, actually gives a ‘machine’ for starting with a $T_3$ space having two points that cannot be separated by a continuous real-valued function and produc...

nope, the construction seems to still not quite compute

sad, I was interested in this "machine" producing $T_3$ strongly connected spaces

think_meaning_buildß

Drop him a reminder in the comments

Jakobian

I don't really remember what the deal with the construction was

I kind of gave up on it because I couldn't make sense out of it

think_meaning_buildß

19:54

@SineoftheTime professors to the young

As in professing or preaching to them words of wisdom.

Jakobian

20:07

@think_meaning_buildß I did. I'll see what his response would be and if nothing comes out of it, I guess I'll try asking on mathoverflow

think_meaning_buildß

Sounds like a good plan.

Pizza

21:19

Hi

think_meaning_buildß

yo wazzup

Bob

Hello

In the typical college Calculus class, are colculators allowed?

on exams, that is

think_meaning_buildß

nope, not typically

Bob

I think that is good

Sine of the Time

it depends, but usually you don't need a calculator to do calculus

"Believe in calculus, not in calculators"- Sun Tzu

Bob

21:28

well, what about students that want to take calculus but have trouble with negative numbers?

Sine of the Time

wdym?

think_meaning_buildß

they need to get over that trouble, fast

Bob

I know of a case where a young lady wanted to take Calculus ( required for the major) but could not do -1 +2(-3) without a calculator

I recommended dropping Calculus and doing a lot of remedial work

think_meaning_buildß

good recommendation

Algebra1 should be next on her list

Pizza

Determine the solutions $y(t)$ for $t > 0$ of the Cauchy problem:

$\begin{cases}
-y''(t) + 3y'(t) - 2y(t) = t\left(e^{-2t} - e^{-3t}\right) \\
y(0) = 0 \\
y'(0) = 0
\end{cases}$

Bob

21:32

thanks think_meaning_build

bye

think_meaning_buildß

cya

Pizza

Does anyone see a way to do this exercise in a simple way?

Sine of the Time

$y=0$ is a solution

since the solution is unique, $y=0$ is the solution of the IVP

Pizza

There is a point where the inverse Laplace transform is calculated but an absurd system comes out

Anti-transformed*

Sine of the Time

Are you sure the text is correct?

Pizza

21:37

Wait

It could be as you say, maybe it was done on purpose

Sine of the Time

it'd be too good to be true

copper.hat

$y=0$ is not a solution.

Ben Steffan

why is $y = 0$ a solution?

yeah

copper.hat

it is not homogenous.

Sine of the Time

why I subsituted $t=0$ instead of $y=0$ ? :(

Ben Steffan

21:39

@SineoftheTime we all make silly little mistakes :)

copper.hat

i make silly big mistakes. i mean uuuuuugghhh.

Sine of the Time

:)

@Pizza you have to solve it using LT?

Ben Steffan

me too. like deciding to study mathematics :))

copper.hat

:-)

Pizza

@SineoftheTime Yes :(...

$-s^2y(s) + sy(0) + y'(0) + 3sy(s) - 3y(0) - 2y(s) \\
= -s^2y(s) + 3sy(s) - 2y(s) \implies y(s)\left[s^2 - 3s + 2\right]$

This should be the first member

copper.hat

21:45

whoa, slow down, you need to take the Laplace transform of both sides.

Pizza

Yep

$\mathcal{L}\left[t e^{-2t}\right] = \frac{1}{(s+2)^2} \quad \mathcal{L}\left[t e^{-3t}\right] = \frac{1}{(s+3)^2}$

This

So $y(s)\left[s^2 - 3s + 2\right] = \frac{1}{(s+2)^2} - \frac{1}{(s+3)^2}$

$y(s) = \frac{1}{(s-1)(s-2)(s+3)^2} - \frac{1}{(s-1)(s-2)(s+2)^2}$

copper.hat

some partial fractions/residues and you are done.

Pizza

@copper.hat Now I have to use the simple fraction method

copper.hat

i don't know what that is

Pizza

$\frac{A}{s+3} + \frac{B}{(s+3)^2} + \frac{C}{s-2} + \frac{D}{s-1} - \frac{E}{s+2} - \frac{F}{(s+2)^2} + \frac{G}{s-2} + \frac{H}{s-1}$

Shouldn't I do that?

copper.hat

21:52

that woudl be partial pfractions

different name same stuff

yes, do that.

Pizza

Oh yes sorry I wrote it wrong

@copper.hat Ok so I calculate the inverse transform for each term?

copper.hat

well, figure out the $A,B,...$ and then the answer is the sum of the inverses. you may want to factor to make it look nice:-)

Pizza

I got this

$y(t) = \\ Ae^{-3t} + Bte^{-3t} + Ce^{2t} + De^t - Ee^{-2t} + Fte^{-2t} + Ge^{2t} + He^t$

copper.hat

i hate doing such computations, i use macsyma

Sine of the Time

Are you allowed to use tables of inverse LT?

Pizza

21:58

@SineoftheTime I don't know ...

But anyway, for now I don't feel ready

@copper.hat I made some mistakes with the symbols above, I corrected in this last line

Sine of the Time

Did you find the coefficients?

Pizza

I'm not sure what to do now

At this point

Sine of the Time

you can find C,D,G,H using cover up method

$y(s)=\dots$ then $(s-1)y(s)|_{s=1}=H$

or no?

Pizza

I think I can

copper.hat

compute the values of the A,B with the $s$ expressions. you know $y(s)$. for example, consider $(s+3)^2y(s)$ and take the limit $s \to -3$. that will give $B$. its a grind

Pizza

22:11

@copper.hat Even as Sine say would be correct yes?

copper.hat

well, no, you need to eliminate the higher powers first

since $1$ is a simple pole then you can do what Sine suggests for the $1$ pole.

Pizza

Oh okay

But could a system have been created here?

Sine of the Time

In the case $s+2$ and $s+3$ you can't, but why $s-1$?

copper.hat

$${{9}\over{400\,\left(s+3\right)}}+{{1}\over{20\,\left(s+3\right)^2
}}-{{7}\over{144\,\left(s+2\right)}}-{{1}\over{12\,\left(s+2\right)^
2}}+{{7}\over{144\,\left(s-1\right)}}-{{9}\over{400\,\left(s-2
\right)}}$$

i think, i could have made a mistake...

@Pizza not sure what you are asking

Pizza

@SineoftheTime I intend to make the system the same as in integrals with simple fractions

@copper.hat i mean this system ^

Sine of the Time

22:17

yeah you can but it'll take you half an hour to solve it

Pizza

@copper.hat Is this the residue method?

@SineoftheTime Oh yes indeed I was looking for the fastest method

copper.hat

well, it is partial fractions

Sine of the Time

copper's strategy is faster

I'm pretty sure we've discussed the "cover up" method a couple of months ago

copper.hat

take the expression you have above where you wrote Shouldn't I do that.
combine the expression so that the denominator is the same as $y(s)$'s denominator. now match the constant and $s^k$ terms to get the various values.

you can figure out $D+H$ more easily.

multiply bothsides of $y(s) = ...$ by $s-1$ and take the limit as $s \to 1$.

same for $B$ and $F$.

the others you need to grind through

there is no magic here, just things to try

Sine of the Time

if you have to use LT, then there's no shortcut

just brute force

Pizza

22:28

Mm ok

@copper.hat So let's assume this is correct

copper.hat

assume what is correct?

Pizza

$y(s) = \frac{9}{400(s+3)} + \frac{1}{20(s+3)^2} - \frac{7}{144(s+2)} - \frac{1}{12(s+2)^2} + \frac{7}{144(s-1)} - \frac{9}{400(s-2)}.$

copper.hat

that would be cheating, i added it as way of checking your work

once you have the coefficients then you have the solution

Pizza

Ok if i understand ,i think you did it like this: $E = \lim_{s \to 1} (s-1)y(s)$

$(s-1)y(s) = \frac{-2s - 5}{(s-2)(s+3)^2(s+2)^2}$

copper.hat

no, that is how i computed $D+H$ the terms with $s-1$ in the denominator

Pizza

22:34

$E = \frac{-2(1) - 5}{((1)-2)((1)+3)^2((1)+2)^2}$

copper.hat

i'm not going to do the arithmetic :-)

Pizza

$E = \frac{-7}{(-1)(16)(9)} = \frac{-7}{-144} = \frac{7}{144}$

@copper.hat ah

Sine of the Time

@Pizza here you only need 6 constants

you can group D+H and C+G

copper.hat

there is a disconnect. the answer there is for $D+H$ not $E$.

Pizza

I got confused, I'll look better tomorrow. Thanks a lot for the help anyway

Anyway, do you know if there is a way to check if the solution I find for the problem is correct?

copper.hat

22:47

well, i am not guarantying correctness, but if your results match with mine above then it seems likely that they are correct?

also, you can check that the solution satisfies the ODE and initial conditions

Pizza

Maybe wolfram too? I haven't checked yet

Last time it gave me strange result

copper.hat

i think just actually doing the computations and double checking will be faster

Pizza

23:04

👍

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$Mathematics$ Mathematics