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6:39 AM
@Knight Yea f(|x|) is continuous at every x∈[−2,2]. Not sure why you wrote "0 < x < 2", since it is supposed to apply for the whole domain.
 
 
2 hours later…
8:09 AM
Isn't this wrong? Or did I forget something about integration :D
Shouldn't it be ln | ... |?
 
8:31 AM
Asking since I tried it two ways: u = k - T and v = T - k. Since I have no clue about T or T_0 I'm not sure this integral is even possible without assuming something like 'T_0 - T' is positive.
 
@Threnody I personally never put absolute value signs because they are wrong if we want to deal with anti-derivatives or path integrals in the complex plane.
If you think about it, removing the absolute value does not make the integral wrong even for real integrals, as long as the integral is over an interval that it can be integrated in the first place (and ln is defined by a suitable branch-cut).
 
Yes I agree - so in the example I linked I'm supposed to assume T_0 - T is positive, right?
 
As for your inquiry about assuming T0−T > 0, that is a good question, and I'm glad you spotted that issue. I have complained many times about people not being rigorous in solving differential equations and division by zero is one of my frequent complaints.
Let's use x instead of T so that it's not mixed up with t.
We have dx/dt = k·(c−x) for variable x varying with parameter t, for some constants k,c.
 
Ironically this is supposed to be easy for me but this threw me off for a second :)
Yes - I follow
 
We cannot divide by (c−x) just like that.
 
8:42 AM
I agree.
 
It is actually a rather complicated business to solve the differential equation rigorously. You can check out the following post (linked from my profile under "Common oversights"):
9
A: Whence the "everything is linear" phenomenon, and what can we do about it?

user21820I've read all the existing answers long ago but still feel that none have gotten to the heart of the issue. We obtain mathematical results through a process of reasoning. That reasoning must be logical and enough to convince anyone that our results are correct given our initial assumptions. That ...

Skip the first half and search for "terrible example of sloppiness".
The example there is different from the one you have here, because that example unambiguously illustrate that you can get wrong answers by doing the invalid division. So it's not that an invalid technique can still get the right answer; it may not!
But the key techniques are all the same.
In particular, the key step for the example you have here is that you can start from any time t where x≠c, and then extend to a maximal time interval around t where x≠c. This needs the supremum axiom. If you need technical details for this, let me know, but I hope you can try it yourself first.
Then on any such interval you can divide by (c−x) to get 1/(c−x)·dx/dt = k and hence ∫ 1/(c−x)·dx/dt dt = ∫ k dt + p, for some constant p. Then using the substitution theorem (conditions are that dx/dt and ∫ 1/(c−x) dx are both defined), we get ∫ 1/(c−x) dx = k·t + p.
Note that the constant p applies to that interval alone! So if you couldn't extend from your starting point t to the whole interval you are interested in then you may have a different constant in each piece! This is why the solution in the linked post really gets 3 separate pieces!
In the example here you are lucky because after finding the solution for each interval you can prove that the maximal interval around t where x≠c extends to the whole interval of interest.
 
9:01 AM
@user21820 By 'extend' here you mean we're trying to find the biggest possible interval for which the result holds when we integrate assuming x!=c, with the interval centered at some t? I'm getting lost in the language.
 
@Threnody Start with any t0 in the domain of interest. Find the largest interval (a,b) such that t0∈(a,b) and whenever t∈(a,b) you have x≠c.
Then solve the differential equation for (a,b) because you can divide by (x−c) everywhere in that interval.
So that the mathematical equations I wrote above hold.
I don't mean you have to explicitly find a,b at the start. You may not be able to. But by the supremum axiom you can define a,b to be like that.
After solving the differential equation on (a,b), you can then look at the solution to figure out what a,b are.
 
@user21820 Yes - I can understand that... I tried the example you linked to in your answer and got similar solutions except for 1
I don't understand why in the non-trivial solutions, you have a clause for x<=a
Or rather - I don't understand where the 0 comes from.
 
If your solution is not exactly the same as mine, then it's wrong.
You really must do the interval extension correctly.
For the example in the linked post, if you start from a point where y≠0, your interval may not extend to the whole domain.
And in fact you will find that in one direction you can only extend it until some finite point.
So by some judicious reasoning you will be able to figure out that you must have 3 pieces.
Oh sorry that example in the linked post has 2 pieces, not 3.
I mixed up with another example I used before.
@Threnody: To give you a bit more hints: The solution says within the maximal interval (a,b) we have y = (x−c)^2 for some constant c. But dy/dx ≥ 0, so the solution must stop at c. Thus a = c!
Then you have the part before a to deal with. It cannot be that y≠0 when x<a, otherwise it would contradict the part after a that you already solved.
So the part before a must be all zero.
That's the gist. When you fully work out all the details then you know what I mean by "terrible example of sloppiness in most high-school curricula" and "fixing the mistake will require the foundation in logic that most students do not have".
@Threnody: And if you are really bored and want a differential equation whose solution has 3 pieces, let me know and I will dig it up for you!
 
9:30 AM
I'm not sure what my problem is but I have no idea what I'm doing anymore! I should let you know that I've never seen differentiation treated with this 'awareness' of axioms and rigor.

So let me double check:

Case 1: Suppose y is not 0.
We get ∫ dy/dx 1/sqrt(y) dx = ∫ 2 dx
∫ dy 1/sqrt(y) = ∫ 2 dx
This gives us 2.sqrt(y) = 2x + k for some constant k

Case 2:
y is 0.
Then ∫ dy/dx dx = ∫ 0 dx
y = c. (No?)
 
@Threnody Wrong case split.
y is a variable and may be zero at some points and nonzero at other points.
If you want more conventional notation, use function notation.
That is, f'(x) = sqrt(f(x)) for every x∈ℝ.
And do the reasoning properly.
I use variables in Leibniz style because it is possible to make it rigorous without sacrificing the intuitive appeal of that style such as the chain rule.
But for you to see the error now it may be better to use function notation.
 
 
2 hours later…
11:56 AM
1. I stepped away from this for a while. I took a look at your solution in the answer linked previously. I'm going to be honest here but I have little idea what intervals have to do with integration. I do get the intuition. But not the formality.

2. You're picking a point where y is *not* 0. You're proving that IF (subcontext starts here) such a point exists, there must be some at least open neighbourhood centered at that point such that y is not 0 there as well. This is have no idea how to formally prove. I suppose there's a theorem that I forgot that allows me to say this. However I did
I'm also wondering whether it is always the case that 'trivial solutions' are lost when abusing integration this way.
 
12:15 PM
There's also something else that's bothering me a lot. If you refer back to the image I sent a few hours ago, you'll see that the proposed solution can also be -ln(T - T_0) + ... depending on how one changes variables.
Also for better or for worse I asked via email the professor teaching differential eqn's about this (the linked image is from his notes and contains several examples similar to your linked answer as well)
And sorry for the grammatical errors, now that I see them
 
 
2 hours later…
2:20 PM
@user21820: you have given some very important and subtle points in your linked answer. I doubt usual textbooks discuss these while teaching separation of variables for solving differential equations. The example on induction is also nice.
@user21820 : Another variant I learnt was proving the statement $p(n) $ for all $n=2^k$ by proving it for $n=1$ and truth of $p(n) $ implies truth of $p(2n)$. And then combine this with "truth of $p(n) $ implies truth of $p(n-1)$ for all $n>1$" .
 
@ParamanandSingh This variant is commonly taught, but in my opinion it is not a wise idea to teach this to students unless they already have complete facility with standard induction, strong induction and well-ordering (with nested quantifiers in the inductive hypothesis).
In my teaching experience, a huge number of students cannot correctly handle this 'backward' induction technique. They strangely enough can attempt to argue with me that it is illogical lol.
And fortunately, this technique is never needed!
There is a much simpler proof of AM-GM using ordinary induction than the commonly taught one using that funny technique.
Is that an example you're thinking of? =)
 
@user21820: I learnt this from the proof of am-gm
But never used much
 
As I guessed. Well, I have never in my whole life encountered any other situation where it is easier than ordinary induction. If you come across any, let me know! At the moment my intuition doesn't tell me whether there are any such theorems or not.
 
You have more experience than me so i dont think i will find such example.
Your students should consider themselves lucky that there is someone who focuses on these aspects.
But maybe some students may not like such attention to detail and just want to pass your courses.
 
2:39 PM
The simpler proof of AM-GM is via smoothing, which is such a useful technique (whether in continuous or discrete mathematics) that I always teach it. Take any real tuple x[1..k] such that 0 ≤ x[1] ≤ x[2] ≤ ... ≤ x[k]. If all items in x are equal, then AM-GM is trivially true. Otherwise let m be the average of x, and let i,j∈[1..k] such that x[i] < m < x[j]. Then x[i]·x[j] ≤ m·(x[i]+x[j]−m), and hence if we replace x[i],x[j] by m,(x[i]+x[j]−m) we preserve the sum and not decrease the product.
This increases the number of occurrences of m in the tuple x, so it eventually stops. This is the step that uses induction, so if we want to make it rigorous we need to set up the induction hypothesis. If we just want to convince the student of AM-GM, we don't need to do that.
At the end all items in x are equal (to m), and we are done.
@ParamanandSingh Concerning teaching separable differential equations, it's a bit unfortunate that it's in the syllabus at high-school level in many places. And there's nothing much we can do when teaching at that level; the student is simply not ready for real analysis, much less of this sort (that I never even saw in undergraduate courses).
(Actually we don't need 0 ≤ x[1] if we state AM-GM in the exponentiated form x[1]·x[2]·...·x[k] ≤ ((x[1]+x[2]+...+x[k])/k)^k.)
 
@user21820: I knew that proof (it's my favorite) too, but I was not aware that the technique is called smoothing.
And speaking of syllabus, I consider the entire high school math syllabus a big fraud at least in our country. Man do teach us one thing properly without the sentence "the proof is beyond the scope of book/syllabus".
@user21820: If everything is outside the scope then what lies in scope??
 
@ParamanandSingh The general idea is to do a local transformation to make something better, and show that repeating that either terminates in the desired optimal configuration or tends to that in the limit.
You might find it an amusing exercise (because it's not trivial) to show that the greedy algorithm for the change-making problem works for fibonacci denominations. That is, for any natural n the greedy algorithm to express n as a sum of fibonacci numbers by cutting off the largest possible fibonacci number at each step really does achieve the minimal number of terms possible.
 
2:55 PM
The statement of the problem itself sounds interesting. Don't give me any further details. I will try it out and maybe discuss later
 
@ParamanandSingh Hahaha... I believe we can design a good syllabus for high-school mathematics that does everything rigorously enough (even without formal logical reasoning) except for a handful of theorems that are so useful that they should still be taught. This handful includes CLT.
@ParamanandSingh Sure! I look forward to seeing your solution.
 
@user21820: If I can't solve then I can always postmy attempt on math.se. :)
 
Hahaha... I'd prefer if you don't publicize it too much, because I like to have a handful of such problems to test students with.
Not for grading purposes, but for evaluation of knowledge.
 
OK then I will discuss in chat
OK it's gonna be dinner time soon. Will get something to eat. I will catch you later. Bye!!
 
See you!
 
 
2 hours later…
5:31 PM
@user21820 Do you have other 'tricky' induction examples like the one in the answer?
Thanks
 
5:50 PM
@Threnody Yes I do. But first you have to formally prove the induction exercise I gave you over PA. I do not think it's a good idea to try hard problems because generally the harder the problem the harder it is to explain why a wrong proof is wrong, so ability to formalize is crucial.
=)
 
6:07 PM
@user21820 Ok, could you kindly link the exercises again if it's not too much of a pain? It's probably tens of days in the past now. :D
 
PL:
in Logic, Jan 28 '19 at 2:56, by user21820
(1) A or ( B and C ) implies ( A or B ) and ( A or C ).
(2) ( A or B ) and ( A or C ) implies A or ( B and C ).
(3) ( A or B ) and ( B or C ) and ( C or A ) implies ( A and B ) or ( B and C ) or ( C and A ).
(4) ( A implies B ) or ( B implies A ).
(5) ( A implies B or C ) implies ( A implies B ) or ( A implies C ).
FOL:
in Logic, Feb 20 '19 at 6:30, by user21820
In these exercises, S is a type, and P is a property, and Q is a 2-parameter property (i.e. "Q(x,y)" is a statement about "x" and "y").
(Q1) not forall x in S ( P(x) ) implies exists x in S ( not P(x) ).
(Q2) not exists x in S ( P(x) ) implies forall x in S ( not P(x) ).
(Q3) exists x in S ( x in S ) implies exists x in S ( P(x) implies forall y in S ( P(y) ) ).
(Q4) forall x,y,z in S ( x=z and y=z implies x=z ).
(Q5) forall x in S ( forall y in S ( Q(x,y) implies P(x) ) ) iff forall x in S ( exists y in S ( Q(x,y) ) implies P(x) ).
PA:
> ∀k∈N ∃m∈N ( k=m·2 ∨ k=m·2+1 ).
I don't care whether you post your attempt using ascii text or symbols. As long as I can read it easily and copy paste to discuss. =)
And feel free to try whichever you like, but try to aim for the level you are at now. If the PA exercise is too difficult, do the FOL ones first.
 
Thank you :)
 
@user21820: hello again! I think I have found the solution to the fibonacci problem we discussed earlier. I hope it is correct proof
 
@ParamanandSingh Sure, let's see it.
 
Suppose we express the number $n$ as sum of $k$ fibonacci numbers $F_{n_1},F_{n_2},\dots,F_{n_k}$
Then I claim that if $k$ is to be minimal then we can choose all $n_i$ as distinct
If some are equal then we know that $2F_{n_i}=F_{n_i+1}+F_{n_i-2}$ so we can replace two equal terms with two unequal terms
Next there is another obvious claim we can't have $n_i$ consecutive otherwise we can combine two terms to get a single fibonacci
These are obvious and we thus have $n_i$ as strictly increase sequence with $n_{i+1}>n_i+1
 
6:18 PM
@ParamanandSingh Just because you can make a replacement and make two terms unequal doesn't imply that the replacements won't collide with other terms.
 
Let's keep on replacing equals with unequals
 
Same for combining consecutive terms; it may cause duplicate terms.
@ParamanandSingh Then you need to prove that it terminates. It is not at all obvious.
 
Since the number of terms in finite we will end up somewhere
 
No, because you did not show that number of duplications decreased.
 
Ok this is something I did not dwell, please let me get to the final part of my argument and then I will fix this problem
 
6:20 PM
Ok.
 
So assume $n$ as sum of these fibonacci numbers whose indices $n_i$ satisfy minimum gap of $2$.
Then I claim that $F_{n_k+1}>n=\sum_{i=1}^k F_{n_i} $
Subtracting $F_{n_k} $ from both sides gives us $F_{n_k-1}>\sum_{i=1}^{k-1}F_{n_i}$
Since $F_{n_k-1}>F_{n_{k-1}+1}$ the problem is reduced to $k-1$ and induction can work on $k$
I think this can explain that $F_{n_k} $ is the largest fibonacci less than equal to $n$.
There may be some typing mistake, but I think this part should work fine.
 
What "induction can work on k"?
Is it induction of the entire problem, or something else?
Just this inequality?
 
Yes this inequality
 
Ok then this part is fine.
Unfortunately, the earlier part is actually the core difficulty of this problem.
So, take your time to find a solution for that part! =)
 
The problem which you highlighted needs a bit more care. I tried a few simple examples and found that I could terminate the process in finite number of steps. But yes I need proof. Will see if I can manage before sleep (30-45 min left).
Will get back to you again.
 
6:31 PM
I'll need to go off soon, so I'll be back much time later. No hurry.
See you next time! =)
 
Ok fine bye bye
 

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