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Feb
26
awarded  Explainer
Nov
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awarded  Yearling
Sep
11
comment How to show that $y'(x)=y(x+1)$ is not an ODE?
Can you show how to exactly use these to answer the question in the title? Because I still don't see it :p I do now see how $y(x+1)$ is not determined by $x,y(x),\dotsc,y^{(n)}(x)$ but that's not the whole story, is it.
Sep
11
comment How to show that $y'(x)=y(x+1)$ is not an ODE?
@Ian: I'm not very good at analysis, so you have to be more specific. Are you referring to the implicit function theorem? In our definition of ODE, nothing extra is assumed about $F$; it's just a partial function on $\mathbb R^{n+2}$.
Sep
11
comment How to show that $y'(x)=y(x+1)$ is not an ODE?
To use this I guess I would have to find $f$ and $g$ like this which are solutions to the DDE. Then both would be solutions to the proposed ODE $F(x,y(x),\dotsc,y^{(n)}(x))=0$ with the same initial conditions at $x_0$, so $f=g$..? But I don't even have any idea how to prove the existence of these $f$ and $g$. Or is there another way?
Sep
11
comment How to show that $y'(x)=y(x+1)$ is not an ODE?
@Ian: I'm not sure I follow. It seems that you're assuming that every ODE can be expressed in the form $y^{(n)}=f(x,y,y',\dotsc,y^{(n-1)})$, but AFAIK that's not the case. It also seems that even then, the example $z=x^n$ doesn't help because it's not a solution to the given DDE?
Sep
9
asked How to show that $y'(x)=y(x+1)$ is not an ODE?
Sep
6
comment Are these proofs logically equivalent?
Nah, while I'm fine with someone using non-standard definitions, if they keep coming up I will demand a reference for them before continuing.
Sep
6
comment Are these proofs logically equivalent?
No, the LHS has 1 and the RHS has 2, which is no problem as the equation 1/2 = 1 - 1/2 shows :)
Sep
6
revised Are these proofs logically equivalent?
added 7 characters in body
Sep
6
answered Are these proofs logically equivalent?
Aug
19
answered Has the opposite category exactly the same morphisms as the original?
Feb
23
answered True or false? $(X\setminus Y)\cup(Y\setminus Z)\cup(Z\setminus X) = X\cup Y\cup Z$, for any sets $X$, $Y$, $Z$.
Feb
9
comment Does the graph of a continuous function have an empty interior?
I think you're confusing image with graph.
Jan
26
answered Class Transitivity Proof
Jan
14
revised Set theory aspects of category theory
added 11 characters in body
Jan
14
comment Set theory aspects of category theory
@ZhenLin Yeah, I was kind of hoping for the converse, so the assumption of the existence of the universe would sound more acceptable, with a proof like "ZFC consistent -> has model -> has transitive model", but I guess the model given by consistency is not necessarily good enough (well-founded) for the Mostowski collapse.
Jan
14
answered Set theory aspects of category theory
Nov
2
awarded  Yearling
Sep
5
comment Question on Zorn's lemma
dkuper: Bounded chains in $\mathbb Q$ do have an upper bound but may lack a least upper bound.