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age 19
visits member for 1 year, 7 months
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NKSSS attendee


Jan
29
comment Why does $\exists x\,\ x = x$?
@Hurkyl Actually, it seems that this convention may be necessary to the avoid the possibility of an empty model. According to this formulation of the Axiom of Infinity, we must presuppose the existence of $\emptyset$ in order for it to be an element of $\mathbb{N}$.
Jan
28
comment Why does $\exists x\,\ x = x$?
OK - that answers everything, thanks!
Jan
28
comment Why does $\exists x\,\ x = x$?
Thank you for the thorough answer, it does make sense. However, as I learned here, it seems that no convention is necessary. $\exists x\,\ x=x$ must be true, since its negation implies $\forall x\,\ x\neq x$, which is the opposite of an axiom of first-order logic.
Jan
28
comment Why does $\exists x\,\ x = x$?
@AsafKaragila I have found the answer to my specific question directly in the first link, thanks.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I apologize that my question may have not been clear - the empty set's existence can be proven from the axiom of specification and infinity, it's just that Wikipedia asserted that $\omega$ was not needed, and I was asking why.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I found what I was looking for in Andres Caicedo's answer here - all one has to do is negate the existence theorem and prove by contradiction!
Jan
28
comment Why does $\exists x\,\ x = x$?
@AsafKaragila thanks for the links. I'd argue that this isn't a duplicate however. I see how the existence of the empty set follows from the existence of a set, but my question isn't about the former.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I would agree that this would be true if I accept the empty set axiom, but I thought the point was that we didn't need one. Without it, $\exists x\,\ x = x$ wouldn't follow from the axiom $\forall x\,\ x = x$.
Jan
28
comment Why does $\exists x\,\ x = x$?
@CliveNewstead I don't see why "there exists a set $x$" is meaningless, but, even supposing it is, I still don't see why I can accept $\exists x\,\ x=x$.
Aug
8
comment Is the vector cross product only defined for 3D?
I don't understand why we have to prove that a cross product exists for any given $d > r$ - for any set of vectors with size $r$, any matrix which has a column space defined by the basis of this set of vectors is guaranteed to have a rank of at most $r$, and so a null space of $d-r > 0$. Since this null space is always nontrivial, a vector satisfying the criteria for the cross product can always be found.
Oct
29
comment Show that the function $g(x)=x^4+x^3+1$ is one-to-one on $[0,2]$
Oh, sorry, never mind. For some reason I interpreted "factor" as "divide out by." My mistake.
Oct
29
comment Show that the function $g(x)=x^4+x^3+1$ is one-to-one on $[0,2]$
Why are you allowed to factor out $x_1-x_2$ if they're equal?
Oct
27
comment Can you approximate a vector field?
It would be one half, because there is an overlap in the information being used for the averages. But the actual implementation of averaging the is immaterial - you could think of lots of ways to do so. Yes, I was referring to a discrete grid of vectors that represents the vector field - that was what your question was asking - how to show general trends - no?
Oct
26
comment Can you approximate a vector field?
Yes - there would be less information about the field in the image. For example, if you plot some vector field $\vec v$ on a grid with steps $\Delta y$ and $\Delta x$, then after averaging $\vec v'(x, y) = (v(x, y) + v(x+\Delta x, y) + v(x, y+ \Delta y) + v(x-\Delta x, y) + v(x, y- \Delta y))/5$ and plotting a vector on every point on the grid with steps $2\Delta y$ and $2\Delta x$, you'll end up with half the vectors you started with.
Oct
8
comment How do i differentiate this function with e to the x and fraction?
Is this homework?
Aug
28
comment Calculate the area on a sphere of the intersection of two spherical caps
Ah, I understand now.
Aug
28
comment Calculate the area on a sphere of the intersection of two spherical caps
Yes - but by "plane representation" do you mean the projection onto the $xy$-plane? Because then the two circles would look like ellipses (and that's why I asked the question that you answered for me before).
Aug
28
comment Calculate the area on a sphere of the intersection of two spherical caps
I'm having trouble understanding your diagram - Is the shaded region the area that you're trying to find? What is its relation to the intersection of the two spherical caps?
Aug
28
comment Proving XY is perpendicular on :CD
Is this homework?
Aug
27
comment How can I describe the area between two ellipses?
Okay, @Lubin. That makes sense now. And an intersection between a circle and ellipse is guaranteed to be a simple type II region. Thanks, Christian.