Here's something to notice, @Abcd. If we switch the last two rows, it's no longer symmetric, but the determinant is still $0$. And then the vector $(1,\omega,\omega^2)$ becomes an eigenvector with eigenvalue $2a+b\omega+c\omega^2$. Here $\omega = e^{2\pi i/3}$ is the primitive cube root of unity. ... At any rate, now we have eigenvalues $2a+b+c$, $2a+b\omega+c\omega^2$, and $2a+b\omega^2+c\omega$. One of those must equal $0$ because $\det = 0$.
Let $a,b,c \in \mathbb R$ such that no two of them are equal and
satisfy $$\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0 ,$$
then the equation $24ax^2 + 4bx +c=0$ has:
a) atleast one root in $[0,\frac 12]$
b) at least one root in $[-\frac 12, 0)$
c) at least on...
If you've read what I've been typing, you know that $\det A$ is the product of the three eigenvalues. $2a+b+c$ is ONE of those. One of the others could vanish. By changing the order of the rows, I get the nice eigenvalues with cube roots of unity.
If $A _ { 1 } \supseteq A _ { 2 } \supseteq A _ { 3 } \supseteq A _ { 4 } \supseteq \cdots$ are all finite, nonempty sets of real numbers, then the intersection $\bigcap _ { n = 1 } ^ { \infty } A _ { n }$ is finite and nonempty.
(technical caveat: mathematica can't actually plot a line of solutions in that way. so what I did was plot the region for which that expression is within 0.1 of zero.)
Point being: I think the only real solutions you get to that are just $2a=b=c$
@SharathZotis it's finite because it's a subset of $A_1$. If the intersection is empty then for every $x$ there is $n$ such that $x \notin A_n$; in particular, if $A_1 = \{s_1, s_2, \cdots, s_k\}$, then for each $i$ we have $s_i \notin A_{n_i}$ for some $n_i$ depending on $i$. Take the maximum of the $n_i$ and $A_{n_i}$ would be empty, contradiction
In which case I'm back to agreeing with Abcd that $2a+b+c=0$ is the only way for the determinant to vanish, subject to the condition that $a,b,c$ be real and distinct
@Semiclassic: What's bothersome here is that the original matrix (before I switched rows to make it a circulant matrix) is symmetric, and so has three real eigenvalues. So you can't conclude just one of the eigenvalues is zero.
How does one catagorize the parameter space of all parametric lines running from $(0,0)$ and "ending" at $(1,1)$? The restriction is that the lines cannot cross each other
Well, I have the factorization in terms of $\omega$, admittedly, but Mathematica gives $\pm\sqrt{4 a^2 - 2 a b + b^2 - 2 a c - b c + c^2}$ as the roots. And symmetry dictates real eigenvalues, so I'm confused.
OK, so the characteristic polynomial collapses to $t(t^2-\alpha) $, where $\alpha$ is the sum of the principal $2\times 2$ minors, namely $\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_1\lambda_3$.
@TedShifrin can you help me develop a more concise question. I want to know the notation for describing a parameter space of lines that start at $(0,0)$ and "end" at $(1,1)$
@Semiclassic: So it seems that we have no conditions on $a,b,c$ other than $a+b+c=0$, so we have two completely free parameters $a,b$ and the roots of that polynomial are $$\frac{-b\pm\sqrt{(b+3a/2)^2+3a^2/4}}{6a}.$$
So, we were learning that there was an additional requirement, beyond the derivatives of the Cauchy-Riemann equations being equal, that is needed for the derivative to actually exist. I just can't seem to remember exactly what the requirement was.
In addition to the Cauchy-Reimann equations being true? Or some detail in the statement I made, such as "If the derivative exists, all directions of approach must agree"?
One derives the Cauchy-Riemann equations from the latter definition. You take real derivatives - one in the direction of 1, another in the direction of i.
But the Cauchy Riemann equations make perfect sense by themselves. It is a theorem (not terribly hard) that if f satisfies the CR equtions, it is complex differentiable.
In fact, something stronger and surprising is true. The derivative is continuous. In fact, the original function can be complex differentiated infinitely many times. (This is harder.)
@LeakyNun see the most recent post on the Lean question: does it have sets? Can working mathematicians work in Lean using set-level constructions instead of "thinking via type theory"?
Hey guys! I have a super quick question for y'all that's not really worth posting a site question for. For any continuous function, f, is it safe to say that if f(x)=a and f(y)=b where a<b, x<y then there will be some f(z)=c where x<z<y, a<c<b? And if so is there some theorem that states this? Or is it just part of the definition of a continuous function?
Goursat's theorem is about C^1, but there is a whole field of PDE where one thinks of these functions as in L^2 and defined the derivative of L^2 functions "distributionally", i.e. based on the way they act on other functions; that's what I'm most comfortable with, and was what I meant by "weak solution" above
but the theorems in that paper are stronger and more in the direction you were thinking.
which has the consequence of the polynomial going from having one root in [0,1/2] to having two roots (one of the roots outside the interval enters) to having one root (the other root leaves)
@Semiclassic: Here you go. Consider $f_\beta(x) = 12x^2+4\beta x-(1+\beta)$. $f_\beta(0)=-(1+\beta)$, $f_\beta(1/2)=\beta+2$. The intermediate value theorem finishes it. No matter what value of $\beta$ you have, there's a root in $[0,1/2]$. (I just did a few cases: $\beta\ge 0$, $\beta\le -2$, $-1\le\beta<0$, $-2<\beta<-1$.)
But, now, how does one discover this without cheating?
So at least one of $f_\beta(0),f_\beta(1/2)$ must be positive, and since $f_\beta(1/4)=-1/4$ we can conclude by IVT that there's at least one real root in [0,1/2] @TedShifrin
Though I suspect I might need to go full java or something in the future cause the zooming is so user unfriendly that it will not help much on investigating the distribution of different types of irrationals in general
Meanwhile, after so much crazy procrastination, I finally start putting my introduction into the thesis. Already 5 references have to be discarded since they don'
Let $a,b,c \in \mathbb R$ such that no two of them are equal and
satisfy $$\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0 ,$$
then the equation $24ax^2 + 4bx +c=0$ has:
a) atleast one root in $[0,\frac 12]$
b) at least one root in $[-\frac 12, 0)$
c) at least on...
@Abcd: We renamed $2a$ as $a$ for convenience. As you said, you have $a+b+c=0$, so $c=-(a+b)$. Your polynomial is $f(x)=12ax^2+4bx-(a+b)$, and we claim that no matter what values $a$ and $b$ have there is a root in $[0,1/2]$. If $a=0$, note that $1/4$ is a root. Divide through by $a$ and consider $g(x)=12x^2-4\beta x-(1+\beta)$. Claim this always has a root in $[0,1/2]$.
Let $a,b,c \in \mathbb R$ such that no two of them are equal and
satisfy $$\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0 ,$$
then the equation $24ax^2 + 4bx +c=0$ has:
a) atleast one root in $[0,\frac 12]$
b) at least one root in $[-\frac 12, 0)$
c) at least on...
