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12:59 AM
Hi. Solving $3y'=x+2xy$, I got $\ln \left | 1+2y \right |=\frac{x^{2}}{3}+C$ after integrating, and finally arrived at
$$y=\pm \frac{e^{\frac{x^{2}}{3}+C}-1}{2}=\pm\frac{e^{\frac{x^{2}}{3}+C}-1}{2}\mp \frac{1}{2}$$.
I checked my result using WolframAlpha, and the solution it came up with was
$$y=Ce^{\frac{x^{2}}{3}}-\frac{1}{2}$$.

That confuses me. I know that we can eliminate $\frac{e^{C}}{2}$ by choosing a constant $D$ such that $D=\frac{e^{C}}{2}$, and since $D$ can be both positive or negative, the $\pm$ in front of $\frac{e^{\frac{x^{2}}{3}+C}-1}{2}$ can be eliminated as well, but how
I meant:
$$y=\pm \frac{e^{\frac{x^{2}}{3}+C}-1}{2}=\pm\frac{e^{\frac{x^{2}}{3}+C}}{2}\mp \frac{1}{2}$$
 
 
1 hour later…
2:25 AM
The word canonical is just designed to intimidate.
I really don't get the point of indefinite integrals.
 
2:51 AM
Seriously? You don’t believe in antiderivatives?
 
Antiderivatives make sense. Indefinite integrals seem to introduce more ambiguity than necessary.
 
To me indefinite integral = antiderivative … I dunno what you’re thinking of.
 
 
1 hour later…
4:24 AM
What is a "dummy variable" in a definite integration?
May I get a book suggestion on Integration that helps to solve most of the problems in Integration but a bit rigorous ( mostly solving problems is necessary)
 
5:23 AM
The $x$ in $\int_0^1 f(x) dx$ is a dummy (bound) variable. It is dummy in the sense that it can be replaced by any other (non clashing) name.
 
5:53 AM
Beacuse we are gonna substitute the limits we call the variable dummy?
 
are you asking where the word came from? one of the senses of the word "dummy" is something that exists only in name. that might be why "dummy variables" are called that. if it isn't, it would be a good retroactive reason to call them that.
 
No what makes them to be called "dummy"?
 
a real valued function on an interval is just a rule that assigns a real number to every number in that interval. it doesn't come with "variable" name. but it's often useful to have one, even if it doesn't matter what it is.
 
Since it takes different values can it be called an arbitrary constant?
 
int 0...1 f(x) dx = int 0...1 f(t) dt = int 0..1 f(p) dp. the choice of 'variable' is just something that works with the notation, so you can write the rule of the function as part of the integral notation.
the constant of integration is something else. that comes up with computing antiderivatives (indefinite integrals), not in computing definite integrals.
 
6:04 AM
No I am asking can the dummy variable be called arbitrary constant?
Since the values of dummy variable is fixed in a particular variable?
 
no, you wouldn't say that. there's nothing "constant" about the dummy variable. it's a symbol that doesn't stand for any particular value.
it may help to simply familiarize yourself with how the notation is used, without worrying too much about why some people call an ingredient in the notation by the name "dummy variable." you could write an entire calculus book without ever using the phrase.
 
So it's just a name
How could I identify the dummy variable?
 
yeah. there's no deeper meaning to speak of.
 
Most of the book use the term "dummy variable"
 
it's the letter appearing after "d" at the end of the integral.
 
6:08 AM
Suppose I have integral from a to b F(t)dx then the dummy variable is?
 
x, although you probably wouldn't be writing that.
 
ohh
What is t here?
How would a function in t be integrated with respect to x?
 
you tell me. i interpret it to mean, integrate from a to b, the function that assigns to the value x in [a,b] the number F(t).
since F(t) doesn't seem to depend on x, this is a constant function. the result would be F(t) (b - a)
 
Oh
Should the limits of integration always be constants?
 
that's up to you. there's nothing wrong with considering the integral from a to b of f(x) dx as a function of b, if you want.
 
6:12 AM
Oh
 
i'd suggest not immediately making everything variable, if you're new to definite integration. start with functions of one variable only with numbers as the bounds of the integral.
 
But the concept of area won't out
 
not because you 'have' to do it this way, but because it's helpful to study the simple cases before you begin letting other things vary.
 
Are they other definitions for definite integration?
Ohh
May I get a book suggestion on Integration that helps to solve most of the problems in Integration but a bit rigorous ( mostly solving problems is necessary)
@leslietownes
 
i don't know what you mean by 'most of the problems in integration.' if you mean the kind of integrals that will pop up in a first year calculus course, almost any calculus book will do. i learned out of stewart, but it's only one of many possible choices.
as far as what's 'rigorous,' there isn't much going on in calculating integrals in first year calculus beyond the fundamental theorem of calculus, and various theorems about differentiation.
most calculus books are pretty good about including the basics of differentiation (maybe with proofs of the theorems in an appendix). the FTC is where they sometimes wave their hands a bit, but it doesn't matter. you can use the theorem like a black box, without knowing how to prove it.
i seem to recall stewart having complete proofs of things like the sum and product and chain rules of differentiation, maybe in an appendix. for the FTC there may have been some pictures drawn and a little bit of 'trust me on this.'
 
6:32 AM
Thank you~
 
you're welcome!
 
dy/dx=f(x,y) what does this notation mean?
 
y is some function (implicitly, of the 'variable' x) and its derivative with respect to x is some expression in x (the 'variable') and y (the function itself).
 
Implicitly meaning it cannot be completely expressed in terms of x?
Also is tangent to a curve a vector?
 
no, here, "implicitly" meaning, there is no surrounding context to tell us what y is, but we can understand from the fact that "dy/dx" is written there that y must be a function of x.
 
6:39 AM
Because the geometric meaning of the above notation is said to tell the direction at a point P in the domain of the function y
Oh so we don't know what the function is? but certainly we do know it is some function in x?
What does it mean that the number f(x,y) shows the direction at that point?
 
7:34 AM
What is the slope field and direction field?
Are both same
?
Why should partial derivative of y wrt x be continuous in R in Picard's theorem?
 
8:35 AM
0
A: Lee Introduction to smooth manifolds problem 6-4

one potato two potatoBy Problem 6-3, there is a smooth function $\tilde{\delta}:M\to\Bbb R$ such that $\tilde{\delta}|_B\equiv 0$ and $0<\tilde{\delta}(x)<\delta(x)$ on $M\setminus B$. Since $M\setminus B$ is an embedded submanifold of $M$, by smooth approximation theroem on $f|_{M\setminus B}$, there is a smooth map...

I uploaded my solution. Can someone check if this answer makes sense?
 
9:15 AM
How to interpret the domain here
$f: \Bbb{S}^{1} \times \Bbb{R}^{1} \rightarrow \Bbb{S}^{1} \times \Bbb{R}^{1}$
seems like acylinder?
$f$ is also referred to as skew product system?
 
Can there be more integral curves passing through a fixed point No right?
 
9:42 AM
So when solving the differential equation adding the constant of integration gives the general solution?
 
How do people deal with all the kinds of families of sets in measure theory
I can't remember them at all
Rings, algebras, semirings, monotone classes, sigma-rings, sigma-algebras, pi-systems, lambda-systems
And many of them can be described in multiple ways too
 
10:00 AM
I guess there's no other option than to draw weird diagrams
 
 
1 hour later…
11:26 AM
e^-y/dy = (1+x^2)/dx Can I take the reciprocal and integrate it?
 
12:07 PM
It is hard to tell what that is
It appears you are trying to write $\frac{e^{-y}}{\mathrm{d}y}=\frac{1+x^2}{\mathrm{d}x}$
but that really doesn't have any meaning
 
 
1 hour later…
1:31 PM
0
Q: John Lee Problem 6-10

one potato two potatoThe following is John Lee's Introduction to Smooth manifolds Problem 6-10. Suppose $F:N\to M$ is a smooth map that is transverse to an embedded submanifold $X\subset M$, and let $W = F^{-1}(X)$. For each $p\in W$, show that $T_pW = (dF_p)^{-1}(T_{F(p)}X)$. My attempt: Let $\dim M = m,\dim X = k...

 
2:13 PM
I don't know anything about physics but I once heard that the universe is smooth Lorentzian 4-manifold. Is this a consequence of something or just a hypothesis?
 
@onepotatotwopotato I think your question would be better suited for PSE chat
 
@onepotatotwopotato space-time gets modeled in this way, but it's one of many models of stuff in physics as a manifold.
Often, the phase spaces of physical problems can be seen as manifolds, and solutions as geodesics on manifolds.
 
So it's a hypothesis?
 
Depends what you mean by "hypothesis".
Models are how we do physics. They have a certain scope, and going outside of the scope leads to issues (e.g. models for classical mechanics can put stuff on the moon, but the moment you try to figure out quantum-scale stuff, the models are no longer effective).
Similarly, modeling space-time as a 4-manifold has perks in many situations, and in other situations you don't want to bother with that particular model.
But given certain assumptions, it accurately models what we understand to be space-time, empirically speaking.
 
@anak Just like assuming speed of light is constant in theory of relativity.
 
2:22 PM
Speed of light in a vacuum is constant because that's the empirical observation, so far.
 
 
1 hour later…
3:33 PM
@robjohn What do you mean shouldn't the notation have dx in the denominator?
How is the differential equation solved dy/dx=y^2/(1-xy) ?
 
That one in particular is an "exact" ODE, is it not?
 
@JaiSriKrishna It is an abuse of notation, but a common one, so it is usually okay, to write $\frac{\mathrm{d}y}{e^{-y}}=\frac{\mathrm{d}x}{1+x^2}$, but the other way around will only cause confusion.
 
So e^-ydx+(1+x^2)dx=0 How is it solved?
@anak It is But I am unable to solve it
with simple integration :(
 
If you have found it is an exact ODE, you can then just follow the cookie-cutter solution for exact ODEs.
 
How is ydx integrated if y is an unknown function in x?
it is like solving integral dy/y^2=integral dx/(1-xy)
How is it solved?
 
3:52 PM
Oh that's not as simple as I thought it was.
Sorry! Are you expecting something to have the Lambert W in it?
 
I don't know it was in the problem list of "differential equations with applications and historical notes"
I have just started the learn the material
Btw it was a verify question so should I go from the solution and show it satisfies the differential equation?
 
What solution do you have?
 
So e^-ydx+(1+x^2)dx=0 How is it solved without using wrong notations?
The solution given was xy=log y +c
And the author says never trust any printed material and asks the reader to verify the solution!
 
Do they just ask "verify that xy = log y + c is a solution to the differential equation ..."?
If so, then they just want you to test that this solution satisfies it.
 
Btw Are u from India?
Yeah.. Thank you @Anak
 
4:01 PM
I am not from India.
 
Oh OK Sorry
 
Don't apologize. :)
 
So e^-ydx+(1+x^2)dx=0 How is it solved?
Without using the step e^-y/dy = (1+x^2)/dx
Since it's wrong
 
Look at your solution xy = log y + c for a moment.
You want to check it solves the differential equation, so you want to introduce dy/dx somewhere.
To do this, you have to differentiate. Can you differentiate both sides of your solution with respect to x?
 
Yeah
Yes I could
we get xdy/dx + y = 1/y (dy/dx)
I think rearranging it gives the differential equation!
 
4:06 PM
:)
 
So e^-ydx+(1+x^2)dx=0 How is it solved?
Without using the step e^-y/dy = (1+x^2)/dx
So here First I divide the complete equation by dx then by rearrangement I get this step e^-y/dy = (1+x^2)/dx which is against the notation, So how can I not get that step?
Also what does dividing by dx mean?
 
You should not be dividing by dx. What makes you think you should?
 
to get the term y'
I thought of it
But now I realize it doesn't make sense
:(
e^-ydx+(1+x^2)dy=0 *
 
4:26 PM
I'm wondering if someone could please help me with my question on sequences?
Pretty sure constructive proofs will be painful and there's probably a slick trick hiding somewhere
Thank you!
 
@esoteric-elliptic What question?
 
@TedShifrin sorry about that. I was in hurry. \dot{y} and \dot{x} with respect to time. This is known as nonholonomic constraints. I would like to know mathematically why this kind of equations can't be integrated even numerically.
 
What is $\theta$?
A second independent variable? A given function of time?
 
@TedShifrin the heading angle of the robot with respect to time.
 
A given function of time?
 
4:40 PM
A coin is flipped twice. For each flip, a marble is placed into a bag:

If the flip was heads, a red marble is placed in the bag.

If the flip was tails, a blue marble is placed in the bag.

You aren't allowed to observe the coin flips or the two marbles in the bag.

Now, you reach into the bag and randomly take out one of the two marbles, and it is red.

You put it back in. Then, you reach into the bag again.

What is the probability of getting blue marble?
Shouldn't the answer be 1/2?
Why am I wrong?
 
What if both marbles are red?
 
@TedShifrin I'm not sure if the time variable appears in the function but the (x,y,theta) vary with respect to time. The robot's position and location.
 
So you have three dependent variables.
 
yes
 
All unknown. Very underdetermined.
 
4:45 PM
I think (x,y) depend on theta but not vise versa.
 
@TedShifrin I am confused about what you want to say. You get first one red then at second you either get red or blue. I don't think first time getting red will have any effect on probability of getting blue.
 
You know there is at least one red. So out of the four equally likely options RR, RB, BR, BB, you have probability $(1/3)(1)+(2/3)(1/2) = 2/3$ of picking R.
 
@TedShifrin The correct answer was 1/4
 
As I said earlier, Croco, $\tan\theta$ is the slope of the path, but not related to $x$ and $y$ directly.
Oh, the question was blue, not red. I get 1/3 at the moment.
 
I was confused why you were talking about red lol
 
4:53 PM
1/2 is clearly wrong. I think I’m right and the 1/4 is wrong.
 
@TedShifrin if I have a Riemannian submanifold $(S,i^*g)\subset (M,g)$ (with the metric induced by the inclusion into $M$), is it a surprise if we find the geodesic equations for $S$ are identical to those on $M$? I feel like it shouldn't be surprising, and I don't know how it would happen that the induced metric on a submanifold could result in different geodesic equations.
 
OK, I see how they get 1/4.
 
0
Q: Rearranging $\{p_j\}$ and $\{q_j\}$ into one sequence $\{r_j\}$ with $s_n \geqslant 0$ and $s_n \xrightarrow{n\to\infty} 0$

esoteric-ellipticI'm trying to prove Lemma $2$ in this paper. The statement is as follows: Let $\{p_j\}$ be a sequence (possibly finite) of non-negative numbers, and $\{-q_j\}$ be an infinite sequence of negative numbers such that $\sum p_j = \sum q_j \leqslant \pm\infty$, $q_j \to 0$ as $j\to\infty$, and $p_j \...

 
@TedShifrin How?
 
This one!
 
5:00 PM
Compute the conditional probability of getting B on the second given R on the first. The probability of R on the first is 1/2, probability of both happening is 1/8.
Anak … you need totally geodesic submanifold. Think of any general surface sitting in $\Bbb R^3$.
There’s a typo, obviously, esoteric, The p’s and q’s are all nonnegative, so $-\infty$ is wrong.
 
1/8 ?
 
Anyhow, esoteric, I don’t get your question. It will be descriptive, just like the proof that you can rearrange any conditionally convergent series to get anything you want.
I don’t yet see how it’s possible if there are only finitely many p’s or if one converges and the other diverges.
 
@TedShifrin Agreed
 
Try $p_n=1/2^n$ and $q_n=1/n$.
 
@TedShifrin Are you suggesting that as a counterexample or something?
 
5:12 PM
Oh, the sums are EQUAL. Never mind.
 
Yes
Did you check the paper, by the way? Hopefully I'm making no error
 
Now it works. Do separately the case they both diverge and the case they both converge to the same number.
 
No, I’m swamped here.
 
I've attached a picture for reference in any case.
OK, thanks, I'll try to work the two cases separately
 
5:16 PM
Hey can someone help me out with the epsilo-N proof of (sqrt(n)+3)/(n+4). I've got that it converges to 0. where can i take it?
 
5:39 PM
R1R2 R1R1 R1R1 R1B1
Point is R1 R2 R1 B1 makes it 1/4
Let's say we got R1X then we got two possibilities (R1R2) (R1B1)
 
@JaiSriKrishna I've put my solution here
You can write $y$ in terms of $x$ and $\mathrm{W}(x)$
 
6:11 PM
I'm struggling to understand the proof of the Laplacian's Rotational Invariance in 2D.
 
How can I prove that: $$(-1)^k\binom{1/2}k=-\frac1{(2k-1)2^{2k}}\binom{2k}k$$ ?
-1
Q: Proving $(-1)^k\binom{1/2}k=-\frac1{(2k-1)2^{2k}}\binom{2k}k$

Ole PetersenHow can I prove that for a natural number $k$ we have the following? $$(-1)^k\binom{1/2}k=-\frac1{(2k-1)2^{2k}}\binom{2k}k$$ It seems like we use a rule for re-writing which I cannot see.

 
@rb3652 How about the polar coordinates formula for the Laplacian?
@OlePetersen Did you just write down the algebra and clear fractions?
 
I know its the correct result. I just don't know to to get there. I tried.
 
Did you write down the product for $\binom{1/2}k$?
 
6:27 PM
That would be $1/2 * (1/2 -1) * (1/2 - 3)* .. * (1/2 - k) / k!$
 
Isn’t there one term too many in the numerator?
Anyhow, then multiply by $2^k$ ….
 
6:41 PM
If I multiply with 2^k would it simplify the term?
 
It removes the 1/2s, yes. And you do need to stop at $1/2-(k-1)$. You will also need to rewrite $1\cdot 3 \cdot\dots \cdot (2k-1)$ as $(2k)!/2^kk!$, as is standard.
 
7:02 PM
Hey guys. I think it is a stupid question :D but i have to ask it: Can i substitute here like i want to or do I get problems because we integrate over y afterwards?
Here is the picture:
https://ibb.co/NFKbP8r
 
7:33 PM
I get $(-1)^{k} \cdot \frac{1/2 \cdot (1/2 -1) \cdot (1/2 - 2) \cdot ... \cdot (1/2 - (k-1) ) }{k!}
= (-1)^{k} \cdot \frac{1 \cdot (1 -1/2) \cdot (1 - 2/3) \cdot ... \cdot (1 - (k/2-1/2) ) }{2^{k} k!} $
 
7:57 PM
Those are some terrible algebra skills. You’re saying $\frac12-1 = \frac12(1-\frac12)$.
 
$= (-1)^{k} \cdot \frac{1 \cdot (-1) \cdot (-2) \cdot ... \cdot (2(1-k) ) }{2^{k} k!} $
 
You should have $1(-1)(-3)\cdots (1-2(k-1))$ in the numerator.
 
8:15 PM
How can we prove that $1\cdot 3 \cdot\dots \cdot (2k-1) = (2k)!/2^kk!$
 
@TedShifrin Hi. in question 5.2.13 in "Multivariable Mathematics" we have three positive numbers $x,y,z$ such that $xy^2z^3=108$ and we are asked to find the minimum value of the sum $x+y+z$. I did the problem in one way, by setting f:\mathbb{R}^3\to \mathbb{R}, f(y,z)=\frac{108}{y^2z^3}+y+z$ and then from $Df=0$ I got the point $(x,y,z)=(1,2,3)$
I mean, the point $(y,z)=(2,3)$ so $x=1$ and the sum is $6$. Now I have found a compact set containing this point and since the function $f$ is continuous here and this is the only extremum point and outside this compact set the function always attains a higher value we can argue that this is the minimum point, right? Now, I have tried to do the same problem using Lagrange multipliers and I also get the same point but can we argue in the same way to show that this point is the minimum point?
 
Write out some examples, Ole. You need to cancel out all the even numbers.
@lirenzo no, this is not on $\Bbb R^3$.
 
Yes, that was a typo (I was thinking about the three numbers $x,y,z$ so I wrote $\mathbb{R}^3$ even if $f:\mathbb{R}^2\to\mathbb{R}$)
 
Obviously, Lagrange is the way to go, but it doesn’t help you understand why a minimum must exist. If you made the (challenging) argument to find an appropriate compact set, then fine. But you need the right domain to start.
Well, it’s not $\Bbb R^2$, either.
 
8:49 PM
@Feynman_00 This is true only when $V$ is finite dimensional.
 
as a domain, $\Omega=\{(y,z)\in\mathbb{R}^2:y\geq 1, z\geq 1, y+z\leq 10\}$ should work
so, since on the boundary and outside my compact domain the function is always greater than the value I found using Lagrange multipliers, the point $(y,z)=(2,3)$ is the minimum point and $f(2,3)=6$ is the global minimum of the function (with this constraint of course)
 
You shouldn't be using Lagrange multipliers. You should just find the critical point of your $f$ above and evaluate.
I'm not sure how you did the estimates on $0<y\le 1$, $0<z\le 1$, etc.
 
9:09 PM
@TedShifrin well, for the case $0<y\le 1$, $0<z\le 1$ we have that $f(y,z)=\frac{108}{y^2z^3}+y+z>\frac{108}{y^2z^3}>108>6$ and for the other cases I argued similarly that $f(y,z)>6$
I did the problem the way you mentioned, by finding the critical point and using the fact that $f$ is continuous and the domain compact. I was just curious how one could do the same problem using Lagrange multipliers since it is a technique that seems very well suited to a problem such as this one
 
9:34 PM
Be careful. What if $y\le 1$ but $z$ is big?
Oh, well, if $z>6$, we're fine. But it's a bit more delicate than you're doing, I think. @lorenzo
 
10:16 PM
@TedShifrin You are right: I checked the inequalities and some result (in my inequalities) weren't fully justified; I tried by modifying the bounds and after several tries I get the following domain $\Omega=\{(y,z)\in\mathbb{R}^2:y+z\geq 6, \frac{1}{12}\leq y,\frac{1}{4}\leq z\}$ which is such that $f(y,z)>6$ on the boundary and outside of it
Thanks
 
Cool.
 
 
1 hour later…
11:21 PM
I always find Runge's phenomenon fascinating.
 
11:40 PM
i liked it better when it was called the Gibbs phenomenon.
 
11:58 PM
I prefer syzygy.
 

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