There are two common ways of defining $π$, and they are interestingly and non-trivially equivalent.
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If you wish to define $π$ as the ratio of a circle's circumference to diameter, you would first have to define arc length, which is highly non-trivial. ...
Maybe I should post an answer, explaining how really $e^{iπ} + 1 = 0$ is forced by our choice of what we want the exponential function to be. This may either make one more fascinated by the way it is so forced, or less fascinated because now the mystery is more clearly unraveled. $\overset{..}\sm...
@LeakyNun I shall explain how to solve your question generalized to any path-connected domain D. I shall use the framework for differentiation described in the following post, which if you wish you should be able to convert it into an 'ordinary' proof.
The answers so far are arguably incorrect; they merely give sufficient but not necessary conditions, and one of them even states that their conditions are necessary. We do not need differentiability in some (open) neighbourhood of the point, even for the conventional (very restrictive) definition...
Okay let's begin. Your question is actually the motivation for defining exp in the first place! If you have some function exp on C such that exp'(x) = exp(x) for any x in C, then for any variable y varying with parameter x in C such that dy/dx = y at every point, we have (at every point) d(y·exp(−x))/dx = dy/dx·exp(−x) − y·exp(−x) = 0, and hence (by Rolle's theorem equivalent) y·exp(−x) = c for some constant c in C. Thus y = c·exp(x).
Note that this applies to any arbitrary (smoothly) varying parameter x in C, and hence for any path-connected D within C (one of which is R) and function f : D → C such that f'(x) = f(x) for every x in D, we have f(x) = c·exp(x) for every x in D, for some c in C.
@LeakyNun: Sorry I missed out a crucial condition; we need also exp(0) = 1, which is used to show that exp(−x) = 1/exp(x). d(exp(−x)·exp(x)) = −x·exp(−x)·exp(x) + exp(−x)·exp(x) = 0 for every x in C, and hence exp(−x)·exp(x) = exp(−0)·exp(0) = 1. Now the second post I linked shows how to actually obtain function exp, motivated by the successive approximations.
Many textbooks/teachers will first prove termwise differentiation of power series, before applying it to things like this. But exp is special and needs nothing more than direct bounds as I did in that post.
So you can see that solving that kind of differential equation is trivial once you have exp. You can similarly solve any first-order differential equation:
Given any continuous f,g on D within C and variable y varying with parameter x in D such that dy/dx + f(x)·y = g(x), we have that ∫ f(x) dx is defined at every point, and hence d( y · exp( ∫ f(x) dx ) )/dx = ( dy/dx + f(x)·y ) · exp( ∫ f(x) dx ) = g(x) · exp( ∫ f(x) dx ), which gives y · exp( ∫ f(x) dx ) = ∫ g(x) · exp( ∫ f(x) dx ) dx + c for some constant c.
In this framework, with respect to parameter t, ∫ z dw is defined as a variable such that d( ∫ z dw )/dw = z whenever ( Δz≈0 and Δw→0 ) as Δt→0. It could be called an anti-derivative, and its existence can be proven by a generalization of the fundamental theorem of calculus. The advantage of doing this is obvious when z and w are not the parameter t.
if you want the formal statement: find all possible functions $f:\Bbb R \to \Bbb R$ that is differentiable everywhere such that $f'(x) = f(x)$ for all $x \in \Bbb R$
@Silent: On the right you see a starred post for free introductory logic texts. Since you have mathematical background, I suggest you go straight to Hannes' lecture notes.
If you don't mind buying a book, one user here @DavidReed recommended Boolos' "Computability and Logic" and you can see some samples here or ask him about that book.
Let $V$ and $W$ be vector spaces over a common ground-field $F$, not necessarily finite-dimensional.
Let $T : V \to W$ be a linear transformation.
Let $T^* : W^* \to V^*$ be its dual, given by $T^* : w^* \mapsto w^* \circ T$.
Can we compare $\dim \operatorname{im} T$ and $\dim \operatorname{im...
I'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 ...
The example there was dy/dx = 2·sqrt(y).
And it was purposely chosen so that just dividing and solving by 'separating variables' would fail to get the solutions.
You can try looking at the hint first and see if you can solve it yourself. But if you want you could just read the sketch of my solution and an alternative.
@LeakyNun In your specific case, one would have to consider x>0 and x<0 separately. In the general framework, at any point where x≠0 and y≠0 we can find an open region around it where both remain non-zero, because the problem guarantees that y is continuous with respect to x, and on any such open region we have 1/y·dy/dx = 1/x^2 and hence ∫ 1/y dy = ∫ 1/x^2 dx + c for some constant c, giving ln(y) = −1/x+c and hence y = a·exp(−1/x) for some constant a ≠ 0.
Note that at any point on a curve that satisfies the differential equation the entire part of the curve in the quadrant containing that point must satisfy the above relation for some constant a. This a may differ for the other parts of the curve in other quadrants.
I basically have given you the entire solution above. You can extract what you want from it. For instance, if you want to find all function f on R such that x^2·f'(x) = f(x) for every real x, then you're basically saying we cannot have two points of the curve at the same x-coordinate. And continuity of f implies that if there is one point in one quadrant then it must extend throughout that quadrant and the other quadrant opposite the x-axis must not have any point.
And since the constant a can be negative, we just have two parts to f, one for x<0 and one for x>0, if they can be joined together at 0.
And in your case, they cannot, because exp(−1/x) → ∞ as real x → 0−.
@LeakyNun: You only have a solution if you restrict to f : R+ → R or f : R− → R.
So the example in my post is more interesting. =P
@LeakyNun: Do you need more detailed explanation of any step? Such questions are never easy or short to solve rigorously.
@LeakyNun Oh okay I made a mistake. I considered points in the 4 (open) quadrants but not on the x-axis itself.
If a point on the curve is on the x-axis itself, it can be extended arbitrarily along the axis.
It cannot leave the x-axis except at x=0, otherwise it will fall into one of the quadrants.
Since a·exp(−1/x) → 0 as x → 0+, the only way to join at x = 0 is through the origin, which means the real solutions must be of the form f(x) = ( x ≤ 0 ? 0 : a·exp(−1/x) ) for every x in R, for some a in R (including a = 0 to allow the x-axis case). We can check that all these are indeed solutions.
@LeakyNun: So, do I pass now? Thankfully I don't have to take any more exams; still so careless.
Take any consistent formal system S that interprets arithmetic and has a proof verifier. Such as S = PA or S = ZFC or any reasonable foundational system. Let S' = S+¬Con(S). Then S' |− ¬Con(S) and hence by some arithmetical argument S' |− ¬Con(S').
I don't have a precise definition "there is an algorithm deciding whether a sentence is an axiom" or something like that, never delved in this area seriously enough
How would the Godel/Rosser incompleteness theorems look like from a computability viewpoint?
Often people present the incompleteness theorems as concerning arithmetic, but some people such as Scott Aaronson have expressed the opinion that the heart of the incompleteness phenomenon is uncompu...
A is a func(nat,string), right? We can determine whether a string x is in the image of B, by checking how many conjuncts it has, oh it has k? then run A[k] and check if it's equal to each conjunct of x.
@LeakyNun By the way, it's called Craig's trick. Incidentally, we don't actually need to think of this if we go via the computability viewpoint, since the proof of incompleteness is pretty much unchanged, as I stated in "Generalization".
> The proof is the same except that you simply need to construct the program to parallelize all the calls to V.
This could be argued to be a 'more' standard enumeration trick, though it too has a name: dove-tailing.
@AgiHammerthief: Hello there! Feel free to bring anything about logic up for discussion here.
@skullpatrol Some others have asked me about it as well. Unfortunately I have not read it before, partly because I don't know where I can obtain a legal copy to read.
Let $V$ and $W$ be vector spaces over a common ground-field $F$, not necessarily finite-dimensional.
Let $T : V \to W$ be a linear transformation.
Let $T^* : W^* \to V^*$ be its dual, given by $T^* : w^* \mapsto w^* \circ T$.
Can we compare $\dim \operatorname{im} T$ and $\dim \operatorname{im...
