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12:35 AM
@DuncanRamage see your browser history :')
@Allie I think, Ted has found out something more interesting and exciting than this group.
 
@LuckyChouhan I was at work and in incognito mode......
 
1:26 AM
@DuncanRamage Browsing pron at work?!
 
1:46 AM
did he really leave :(
 
2:01 AM
those pron sites really mess with you
whatever a pron site is
 
he is left as an exercise
 
@copper.hat just stay away from the kitty pron.
 
i feel like this is something that already exists, but i don't know what it would be called
 
i've been looking for a mountain lion in a nearby park (Tilden) for decades now, but always seem to miss them...
 
suppose we pick a random point in the standard $2n$-simplex. if we add the first $n$ coordinates, what distribution do we obtain?
i guess this amounts to projecting the standard $2n$-simplex onto its first $n$ coordinates and then asking for the distribution of the 1-norm
 
2:10 AM
by a 2 simplex you mean $(x_1,x_2)$ non negative that sum to one?
 
right
 
i would guess uniform?
 
i'd have said "random sequence of 2n positive reals summing to 1" but i wasn't sure if that's precise enough
 
having spent all of 10s thinking about it, so likely wrong...
 
@copper.hat apparently not. here's a 10^6-sample histogram of the n=5 case:
 
2:15 AM
@copper.hat A mountain lion was seen a few blocks from my house years ago. I am glad I wasn’t there.
 
feels very central-limit theorem, but subject to the constraint that the sum falls in [0,1]
 
@robjohn i was always looking from the relative safety of my mountain bike...
 
lower-n examples are broader but still non-uniform
 
@Semiclassical yep, shows how much 10s of my brain is worth
 
this doesn't mean i know how to analyze it unfortunately
 
@onepotatotwopotato that was funny
 
can someone be right as an exercise?
 
2:44 AM
left & right seem so arbitrary.
 
X4J
3:13 AM
Hey, I am trying to determine wether $\int_1^{\infty} e^xsin(e^x)x^{\frac{-1}{2}} dx$ converges or diverges. I've proved it converges, and now I am interested to prove, as I suspect, that it doesn't converge absolutely. Does someone can think of a function which can help (.i.e by using ratio tests)
 
 
4 hours later…
6:53 AM
@copper.hat are they wingers because they offer chicken wings if you join their party
@LuckyChouhan good for him if true? Surely you're speculating with true intentions to offend someone from "this group". What does it tell about you as a person, I wonder?
 
7:25 AM
Is there a free website where I can fill an online pdf form?
 
 
1 hour later…
8:48 AM
Is this answer of mine right?
0
A: Let $\phi: \mathbb Z_{15} \to \mathbb Z_{15}$ with $\phi(2)=5$. What is $\phi(1)?$

ShaunHint: $2$ is a generator of $\Bbb Z_{15}$. Spoiler: Hint 2: The inverse of $2$ in $U(15)$ is $8$. Spoiler:

Someone suggested I should use the additive structure only. But I remember being taught this technique . . .
in Helpful Commentary, 4 mins ago, by Shaun
Please may I have some feedback on the following?
 
9:07 AM
I think the following is wrong:
2
A: Let $\phi: \mathbb Z_{15} \to \mathbb Z_{15}$ with $\phi(2)=5$. What is $\phi(1)?$

J. W. TannerHint: you have $2\phi(1)=5 =5+15=20$ in $\mathbb Z/15\mathbb Z$

You can't divide by $2$. You have to multiply by $8$.
 
9:51 AM
imvho it is a hint, and whether it does or does not have misleading implications is more in the eye of the beholder than it is written in that post (if you think about it, almost any hint about anything can probably be recharacterized in such a way that it sounds misleading, because anything that falls short of a complete answer relies on the reader to fill a gap)
your saying that "you can't divide by 2, you have to multiply by 8" [in Z/15Z) is more baffling to me than that hint is
maybe you're making some very subtle distinction that would become clearer upon, say, distinguishing in notation between integers and elements of Z/15Z, but nobody seems to be doing that
and you seem to agree that 2 has a multiplicative inverse in Z/15Z that either is, or is represented by, 8
because you say as much in your own answer
i think a lot of the language that people use around Z/nZ is unfortunate, and not what i'd use, but this is not an example of language i would identify as misleading or wrong
"is it or is it not a good idea to think or talk in ways that reference division when dealing with integers" is more a pedagogical debate than a math one
i'd say, in general, it is not a good idea, and the last time i taught modular arithmetic in a classroom setting i discouraged people from writing stuff like that. but the linked post is not a very good example of a situation where it is a bad idea to write stuff like that
(and not only because the linked post nowhere says anything about dividing by 2)
 
I wonder why caps lock is written as CapsLk on the key though there is enough space to write the full.
 
ppl r lazy ig
 
10:51 AM
@Shaun I don’t think this is wrong. It’s the way I did the problem when I first saw it. Certainly, $8$ is the inverse of $2$ mod $15$, and if multiplicative inverses is the section in which this was encountered, then I would like to see multiplication by $8$.
 
quotient space with 1 equivalence class
Isn't that called a "equivalence class of degree 1?"
For a spherical topology with north and south poles being singularities is this still considered contractible?
I don't think you can shrink certain choices of loops across the singularities
well I guess I didn't think of the fact that you could shrink the circles onto the singularities themselves...
 
 
1 hour later…
12:37 PM
@robjohn Thanks :)
@leslietownes Thank you.
 
1:06 PM
@Shaun seems like you're asking about the following equivalence? $$ma\equiv mb \pmod{n} \iff a\equiv b\pmod{n/\gcd(m, n)}$$
in this sense division is allowed in congruences
if $m$ and $n$ are coprime, that is when $m$ is invertible, we even stay in the ring $\mathbb{Z}/n\mathbb{Z}$
Perhaps interesting is the fact that even if $m$ isn't invertible, we are still allowed to divide by it, as long as we change our ring to something different
 
1:45 PM
My lecturer defined reducible a element $z \in \mathbb{Z}$ if $\exists x,y \in \mathbb{Z}$ such that $z=xy$ with $x,y \notin \mathcal{U}(\mathbb{Z})$ (the latter being the set of invertible elements of $\mathbb{Z}$. He then proceeds to say that an element is irreducible if $\nexists x,y \in \mathbb{Z}$ such that $z=xy$ with $x,y \notin \mathcal{U}(\mathbb{Z})$, and says that the latter is equivalent to $\forall x,y \in \mathbb{Z}, z=xy \implies (x \in \mathcal{U}(\mathbb{Z})) \vee (x \in \mathcal{U}(\mathbb{Z}))$.
(i) What "such that" and "with" mean logically in this context? I think that the reducible definition means $\exists x,y\in\mathbb{Z}, (z=xy)\wedge(x\notin \mathcal{U}(\mathbb{Z}))\wedge (y\notin \mathcal{U}(\mathbb{Z}))$. Is this correct? Or is there an implication like $\exists x,y\in\mathbb{Z}, (z=xy)\implies ((x\notin \mathcal{U}(\mathbb{Z}))\wedge (y\notin \mathcal{U}(\mathbb{Z})))$?

(ii) I am not convinced by the negation, should't the definition of irreducible be "not reducible" and so $\neg(\exists x,y \in \mathbb{Z}$ such that $z=xy$ with $x,y \notin \mathcal{U}(\mathbb{Z}))$? If
 
@Jakobian Okay. I see :)
Thanks :)
 
 
1 hour later…
3:01 PM
@ZaWarudo why are the two sentences different?
your lecturer is correct, I'm not sure what your issue is
 
@Jakobian: If I want to negate "$\exists x,y\in\mathbb{Z}$ such that $z=xy$ with $x,y\notin \mathcal{U}(\mathbb{Z})$", shouldn't I have to negate "$z=xy$ with $x,y\notin \mathcal{U}(\mathbb{Z})$" in addition to negating the existential quantifier? While I agree with the equivalence with the implication, I don't understand why it is equivalent to the proposition where we negate only the existential quantifier.
In other words, I don't understand why "$\nexists x,y\in\mathbb{Z}$ such that $z=xy$ with $x,y\notin \mathcal{U}(\mathbb{Z})$" is equivalent to $\neg$($\exists x,y\in\mathbb{Z}$ such that $z=xy$ with $x,y\notin \mathcal{U}(\mathbb{Z})$)
 
@ZaWarudo and whats the meaning of $\not\exists$?
its just $\lnot \exists$
so its equivalent pretty much by definition of that symbol
 
@Jakobian: So the symbol $\nexists x, P(x)$ is just a shorthand to say $\neg(\exists x ,P(x)) \equiv \forall x , \neg P(x)$?
 
3:29 PM
@ZaWarudo $\lnot \exists_x P(x)$, yes
try reading those things in English and see how they make sense instead of jumping to symbolics
 
Indeed, I was convincing myself using some sentences in English. Thanks for the help!
 
4:11 PM
FINALLY! I can start chatting! Are there any specific rules for questions I have in the chat just like the regular site? Because I believe if not, then I an actually get some much needed help. I just don't want to break rules when I'm simply trying to improve in math so I can do better in school is all.
 
@YourLordJoyBoy no other than, keep it civil and so on
 
Hi Jakobian!
 
4:29 PM
Grateful for that. I will do my best to keep things civil. But seriously, I appreciate any help I can get. I just want to finally be free of Algebra 1 you know?
Anywho, I've some questions I'd like to not simply get answered but explained so that I know what I am doing. I just want to know what I'm doing so that when it's officially final time, which is fast approaching, I won't panic but know what to do.
 
I'm trying to prove that if $f$ is twice differentiable in $(a,b)$ and there exists $M\ge 0$ such that $|f''(x)| \le M$ for each $x \in (a,b)$, then $f$ is uniformly continuous. Let $x,y \in (a,b)$ and assume $x<y$. By the mean value theorem, there exists $c_{x,y} \in (x,y)$ such that $|f'(x)-f'(y)|=|x-y|\cdot |f'(c_{x,y})| \le M|x-y|$; this proves that $f'$ is Lipschitz continuous in $(a,b)$.

Hence, $f'$ is bounded by some $K\ge 0$. Using again the mean value theorem $|f(x)-f(y)|=|x-y| \cdot |f'(d_{x,y})| \le K|x-y|$ for some $d_{x,y} \in (x,y)$; so, $f$ is Lipschitz continuous as well an
 
@Jakobian Meant to say this before but thanks a ton for reassuring me, Jakobian. :)
 
@Frieren is $(a, b)$ a bounded interval?
If so then this does work, although there is a typo, supposed to be $|f''(c_{x, y})|$
 
@Jakobian: It is not specified. This is the text of the exercise:
 
Do you know of a simple example of a function which would satisfy this on $\mathbb{R}$ but wouldn't be uniformly continuous?
 
4:40 PM
Should I send one problem at a time or can I just ask all of them in a single message? For you guys I'm pretty sure they are not that complicated. But the amount of exponents and variables is messing with my head when I see the terms.
 
@Jakobian oh sure, thanks for noticing
@Jakobian I would say $f:\mathbb{R}\to\mathbb{R}$ defined as $f(x)=x^2$. We have that $f$ is twice differentiable on $\mathbb{R}$, we have $f''(x)=2$ for each $x\in\mathbb{R}$ so $|f''(x)| \le 2$ for each $x \in \mathbb{R}$ but $f$ is not uniformly continuous on $\mathbb{R}$.
 
Yes, perfect, same example as I had in mind
so the author is implicitly assuming $a, b$ to be real numbers
 
So you think that the text of the exercise is a little imprecise?
I always assumed that $(a,b)$ as bounded, if is not specified something like $-\infty \le a \le b \le +\infty$.
 
Its definitely not precise
 
Ok, thanks for checking my work and for letting me think more in depth about the problem:)
 
4:46 PM
@YourLordJoyBoy You should definitely be reasonable about it
I don't think its really a "yes" or "no" type of question, or em, "A" or "B"
 
@Jakobian I certainly want to be. The first one is Perform the indicated operations (14t^4 - 4t^2 + 8t) + (4t^4 - 8t + 27) - (8t^4 + t^2 + 14) And it is in parentheses because it's like that on the actual study guide. I've never understood why they insist on this.
 
there's a matter of not overwhelming the people that could potentially answer your questions as well, homework dumping is not really advised either
if you want to get any answer at all, you have to pay in mind its people that will answer you, potentially
 
Alright. The main point is understanding what I'm doing without the work overwhelming me.
 
there's a lot of things that can be boiled down to single "be reasonable"
 
The work is done so quickly in class that it's hard to keep up, not to mention only having an hour a week for tutoring. And only on one specific day too. Usually I'm good at keeping up, but when it comes to math I tend to be slow on the uptake.
Honestly there are two of these that give me the most trouble. If I can understand what I'm doing on them, I should be able to move on without it being a problem.
I um....hope my question doesn't come off as overwhelming though. I never even considered the question would overwhelm those trying to help.
 
5:05 PM
anyone recognize this version of the letter E?
it doesn't match either $\mathcal{E}$ or $\mathscr{E}$
test: $\mathfrak{E}$
yeah, no good
 
@Jakobian I suppose I should ask, whats the best way to be reasonable about it?
 
@Semiclassical URW Chancery?
 
looks right
 
I'm sorry, I'm lost on that E.
 
@YourLordJoyBoy it boils down to personal judgement really
not sure what you're asking me
@YourLordJoyBoy when I said overwhelming I meant, overwhelming someone while asking multiple questions
@YourLordJoyBoy you don't understand why there's parentheses there?
for addition its not important but for subtraction it is
 
5:19 PM
@Jakobian I don't understand why they wanted to put each problem in parentheses when people usually take them out. For whatever reason on the study guide they're used both for addition and subtraction.
 
They want you to understand how parentheses work perhaps, I don't know
I'm not a teacher and I won't lie to you here
 
No worries there, I'm not expecting that of you.
I simply want to be able to solve these polynomials without the issues I'm having. XD
 
To solve a polynomial, if anything at all, seems to mean solving for its roots
here you're not solving them, but simplifying an expression involving polynomials
this also might be why they wanted to use parentheses, they want you to add first polynomial to second and subtract the third, and want you to think about it this way
 
Yes, that's what I'm trying to understand how to do. But the number of exponents and variables comes off as intimidating for some reason.
 
are you on computer right now? If so, see right upper corner, particularly the part that says LaTeX in chat
we don't have LaTeX here like on the site, you need to add a bookmark to your browser
 
5:26 PM
@YourLordJoyBoy Add one by one then, first write the 14t^4-4t^2+8t then add 4t^4 to it, next add -8t to it and so on
 
@Jakobian I am, actually.
@SoumikMukherjee Alright, I'm on it.
 
@YourLordJoyBoy setting up LaTeX would be more problematic on a phone, its alright if you are
 
In all honesty, never set up LaTeX before.
 
there's also some explanation of it in the link but you would have to go to "computer mode" on your browser to even see the link, and then its also problematic to put it on
 
it's simpler than you might think, at least for chat-level uses
mostly just requires enclosing math in $'s
 
5:28 PM
Wait, I see the link now. Lemme give that a shot.
 
@YourLordJoyBoy well you're not going to download any programs
 
e.g. $14t^4-4t^2+8t$ for $14t^4-4t^2+8t$
 
you can download programs to create LaTeX documents on your personal computer, but that's not necessary here
 
once you can see the latter convert to math, then you've got the start of it
 
its just a matter of adding a bookmark and clicking on it
 
5:29 PM
Ah!
 
something that bothers me is that 50% of the time on the main site the Tex doesn't render automatically and I have to manually render it
but it is better practice for typing in Tex
 
Oi, I'm trying to drag it and it's not doing anything.
 
did you click show bookmarks?
 
do you have the bookmark on your bookmark bar?
 
Wait I think it might have worked
 
5:34 PM
be sure to have "start ChatJax" and not the other bookmark
this way you will only have to click at it once per refreshment of the tab
 
@SoumikMukherjee So you said to write 14t^4 - the 4t^2 + the 8t right?
@Jakobian On it.
How do I test it though?
 
$\LaTeX$
if this renders for you then it works
 
Nope, it didn't.
 
@YourLordJoyBoy yes and add accordingly
 
I have a basic question. How can I verify that the function $\frac1{x}\sin \left(\frac1{x}\right)+\cos\left(\frac1{x}\right)$ is not Lebesgue integrable over $[0,1]$? Graphing it, we see that it oscillates between positive and negative infinity near $0$, but why would the integral of the positive or/and the negative part diverge? This I don't see.
 
5:43 PM
@psie oh this is a pretty cool question because I believe its Kurzweil integrable
 
yummy!
 
This should boil down to some simple bound like $|...|\geq \frac{1}{2x}-1$ for $x$ lying in a significant part of $[0, 1]$
also I know the question doesn't ask for this, but this is Kurzweil integrable because $... = (x \cos(1/x))'$ where we understand that the function $x\cos(1/x)$ is $0$ for $x = 0$
and the integral is $\int_0^1 ((1/x)\sin(1/x)+\cos(1/x))dx = \cos(1)$
curiously, wolfram alpha also lists this as a value of this integral even though this function is not even Lebesgue integrable. Perhaps wolfram alpha is a secret fan of the Kurzweil integrals?
oh... well, or improper Riemann integrals since this is also an improper Riemann integral with value $\cos(1)$
@psie that it oscillates between those, in some sense (it doesn't really oscillate between those values as it never reaches them), is not that important, it does matter that its $\geq \frac{1}{2x}$ on a significant part of the interval $[0, 1]$
 
5:59 PM
ok, how can one specify "significant part of $[0,1]$"?
 
Just solve $|\sin(1/x)|\geq 1/2$
this will give you that area I'm talking about
 
@SoumikMukherjee A negative 8t added to a negative 4t^2
plus a positive t^2
Just making sure I'm doing this right
 
How are you getting the positive t^2?
 
@SoumikMukherjee There was already a positive t^2, I'm trying to bring all like terms together.
For the first set I ended up with 10t^4
The next set of like terms appears to be a -4t^2 along with a second -4t^2 and finally the previously mentioned t^2.
And I was wrong, the second -4t^2 is actually a negative 8t
So the next set is simply, -4t^2 - 8t + t^2
So what are your thoughts, @SoumikMukherjee? Have I screwed up?
 
6:27 PM
@YourLordJoyBoy Okay
 
@SoumikMukherjee I am correct, the answer should be 10t^4 - 3t^2 - 8t + 41.
If I am correct
 
No, you need to to take care of those -ve signs
10t^4 is ok, try the other terms carefully
 
The next one is -4t^2 + t^2 which I believe = -3t^2 and then we have a negative 8t
 
@YourLordJoyBoy what browser are you using?
 
@robjohn Chrome
 
6:34 PM
@YourLordJoyBoy Take a look if its actually +t^2 or -t^2
 
Did you read the later stuff, especially the stuff about Chrome?
 
@SoumikMukherjee Its + t^2
@robjohn Using Chrome on computer rather than on the phone
 
@YourLordJoyBoy What is -(a+b)? a+b or -a-b
 
I believe a+b?
 
6:41 PM
@SoumikMukherjee Yup that figures. Usually if I'm taking a guess, I'm wrong. But this makes sense, since the negative sign is outside the parentheses.
 
7:06 PM
Would anyone mind checking if this looks alright? I'm trying to show that $\frac1{x}\sin \left(\frac1{x}\right)+\cos\left(\frac1{x}\right)$ is not Lebesgue integrable over $[0,1]$.
To that end, for $n=1,2,\dots$, let $a_n=1/(2n\pi + \pi/2)$, $b_n=1/(2n\pi + \pi/4)$. On the intervals $[a_n,b_n]$, which are disjoint, $\sin(1/x)\ge 1/\sqrt 2$ and $\cos(1/x)\geq 0$, and all these intervals are also subsets of $[0,1]$.
So \begin{align}\int_{0}^{1}\left|\frac{1}{x}\sin(1/x)+\cos(1/x)\right|dx&\geq\sum_{n=1}^{\infty}\int_{a_{n}}^{b_{n}}\left|\frac{1}{x}\sin(1/x)+\cos(1/x)\right|dx \tag1 \\ &\geq\sum_{n=1}^{\infty}\int_{a_{n}}^{b_{n}} \left(\frac{1}{b_n}\cdot \frac{1}{\sqrt2}\right)dx \tag2 \\ &= \sum_{n=1}^{\infty}\frac{b_n-a_n}{\sqrt{2}b_n}\tag3 \\ &= \frac{1}{\sqrt 2}\sum_{n=1}^{\infty} \frac{\pi/4}{2\pi n+\pi/2}.\tag4\end{align}
If I did everything right, I think the last series diverges by LCT with harmonic series.
In $(1)$, we used the fact that all the intervals are subsets of $[0,1]$. In $(2)$, we used reverse triangle inequality and the bounds on $\sin$ and $\cos$ listed above, plus the bound $1/x\geq 1/b_n$. The rest is just simplifying.
EDIT: in $(2)$ we did not use reverse triangle inequality at all. We only used the bounds specified.
 
7:29 PM
@YourLordJoyBoy does Chrome allow you to drag links to the bookmark bar as in math.ucla.edu/~robjohn/math/Drag_Bookmark.mp4
 
7:50 PM
@robjohn Yup, and I did
 
and using the link in the bookmark bar did nothing?
 
@robjohn Strangely enough, nothing
 
Hmm, I know that others who use Chrome use ChatJax and it works.
There might be a browser setting that prevents javascript bookmarks from working
I'm not sure; I don't use Chrome
 
Strange. It does not appear to be blocked.
@SoumikMukherjee Hope I'm not bugging you but what now? Whats next?
 
8:10 PM
@psie i have checked the algebra (or verified your claimed inequalities) on a line-by-line basis, but that type of approach should work, so if it checks out line-by-line it looks good to me
 
thank you very much Leslie!
 
@YourLordJoyBoy Keep adding
 
psie: pretty big typo on my part there, meant to say that i have not checked the algebra on a line-by-line basis
but, you know, the logic of it looks solid
 
@SoumikMukherjee Lemme see, like I said before the negative 4t^2 plus t^2 minus the 8t
 
@leslietownes ok, the approach is what counts :)
 
8:22 PM
I am still laughing at this unexpected plot twist
 
@SoumikMukherjee I'm just so ready to be done with Algebra 1. It's been driving me insane.
 
@YourLordJoyBoy Don't worry, it will take time to get used to to this things. Just add things variable wise.
 
@SoumikMukherjee Workin on it. I'm just trying to figure out what I can and can't combine. I don't think the -8t can be combined with the -t^2 and the -4t^2 right? Or can it?
 
It can't
 
@SoumikMukherjee Which would leave me with 10t^4 -3t^ - 8t unless I calculated incorrectly.
And that means I still have 8t plus 27 plus 14
 
8:31 PM
@YourLordJoyBoy No
 
@SoumikMukherjee Where'd I go wrong in my calculations?
 
Slice $S^2$ along the equatorial homology cycle and rotate the top hemisphere by some discrete rotation $\theta$, then glue the hemispheres back together. Does this have a group theoretic description involving $SO(3)$?
 
@YourLordJoyBoy It was -4t^2, not +4t^2
 
@SoumikMukherjee Correct. Thats how I ended up with the -3t^2 since it added to the negative, which is the same as subtracting in that case.
 
What is -4-1?
 
8:35 PM
@SoumikMukherjee One problem. The t^2 is a positive 1t^2.
Otherwise it would be -5t^2
Right?
 
@YourLordJoyBoy The t^2 is not positive, there is a -ve sign outside
 
Are you saying when it left the parentheses it changed signs?
Wait, I get it now. Yes! I forgot about the outside negative sign.
Oh man, of course it was wrong!
That makes it 10t^4 - 5t^2 - 8t + 41? Or should the 8ts be cancling out?
Making it 10t^4 - 5t^2 + 41?
 
@YourLordJoyBoy 27-14 instead of 27+14
 
@SoumikMukherjee Yes yes you're right! So are we saying its 10t^4-5t^2-13?
 
Please don't star every single message, it decreases the aesthetic beauty of the starred messages.
@YourLordJoyBoy +13
 
8:49 PM
@SoumikMukherjee Alright, will make sure not to in the future. And thank you so much! :D
Perhaps we can tackle one more problem?
 
9:08 PM
There's one before it but this I believe is the toughest of them, at least for myself. (-10t^5 + 9t^4 - 3t^3 + 2t + 11) + (-6 - 3t - 3t^2 + 3t^3 + 3t^5)
 
9:42 PM
O.O
 
10:09 PM
@robjohn perhaps you don't mind helping me with this problem?
 
@YourLordJoyBoy Are you talking about getting ChatJax working?
or the math problem?
 
@robjohn The math problem
If its alright with you.
 
Are you looking for roots of -7t^5+9t^4-3t^2-t+5?
 
@robjohn Honestly I don't know
I do know they asked for the polynomial to be added though.
 
Okay, so I already did that. I simply added all terms with the same power of t
 
10:17 PM
@robjohn So that's what you got? I guess I was trying to figure out how to do it. Since that's my overall goal, to understand what I'm doing with each of these problems. Since very likely this problem though with different numbers will be on the final.
 
yes. The t^5 terms have -10 and +3 as coefficients, so you get -7t^5
 
there is a t^4 term in the first and none in the second, so you get 9t^4
the coefficients of t^3 are -3 and +3, so they cancel
etc
 
@robjohn THANK YOU :D
 
You're welcome
 
10:22 PM
@robjohn Just clarifying but you're saying it ends up being -7^5 + 9t^4 correct?
 
and terms with lower powers of t, yes
10 mins ago, by robjohn
Are you looking for roots of -7t^5+9t^4-3t^2-t+5?
It doesn't matter what order you write the terms
 
Okay, okay. So that means and I quote, -7^5+9t^4 +2t? What about the 11 and -6?
@robjohn Yup, thats one thing I do remember.
 
from the comment I just quoted: -7t^5+9t^4-3t^2-t+5
@YourLordJoyBoy that would sum to 5
 
@robjohn And I see where the t comes in now too, a negative t since we added 2 to the 3t. Alright I got it now :D
 
 
1 hour later…
11:34 PM
I forgot how to do trig subs
$$\int_0^t\frac{\frac{B_0^2\ell^2}{R}}{\left(\frac{B_0^2\ell^2}{R}-\frac{m}{t}\right)^2}dt$$ what do
wait this isn't a trig sub
 
11:49 PM
"Trig subs" are no different from any other change of variables...
 
yeah I just did a simple change of variables
 
t is overworked in that expression :( i hope it is receiving overtime pay
 

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