Suppose D in L( P_3 ,P_2 ) Is the diff map defined by Dp= p' , find a basis for P_3 and P_2 such that the matrix of with respect to these basis is [ matrix ]
Suppose $T \in L(P_3(R), P_2(R))$ is the differentiation map defined by $Tp = p'$. Find a basis of $P_3(R)$ and a basis of $P_2(R)$ such that the matrix of T with respect to these bases is
$ \left( \begin{array}{ccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 \end{array} \right)$
I said th...
2 Suppose D G C{V?>( R), V 2 (R)) is the differentiation map defined by
Dp — p '. Find a basis of V 3 (R) and a basis of 7^2 (R) such that the
matrix of D with respect to these bases is
/ 1 0 0 0 \
0 10 0.
\ 0 0 1 0 /
[Compare the exercise above to Example 3.34.
The next exercise generalizes the exercise above.]
3 Suppose V and W are finite-dimensional and T G C(V, W). Prove
that there exist a basis of V and a basis of W such that with respect to
these bases, all entries of M(T ) are 0 except that the entries in row j ,
3.34 Example Suppose D e C(V 3 { R), V 2 (R)) is the differentiation map
defined by Dp — p'. Find the matrix of D with respect to the standard bases
ofV 3 (R) andP 2 (R).
Solution Because (x n )' = nx n ~ l , the matrix of T with respect to the
standard bases is the 3-by-4 matrix below:
0 10 0
M(D ) = 0 0 2 0
\ 0 0 0 3
with regards to a matrix of transformation, I know that if I have a linear transformation from R^n to R^n then the linear transformation of a vector is equal to the (matrix of transformation)vector
It works for every basis. You have to have the coordinate vector of the vector with respect to the basis. If you have my blue book, I can refer you to the page.
@TedShifrin As a last parenthetical, my 14 coordinates are the columns of the matrix here: i.sstatic.net/EY1ku.png. Not terribly revealing at a glance, but there is some symmetry going on.
@KasmirKhaan my group rep professor said that vector spaces that are isomorphic should still be treated differently, for they have different "personalities"
i mean mine didn't restrict to the finite case but 99% of the time the modification from finite to $\sigma$-finite is either easy technicalities or obviously false
@Leaky not liking bourbaki isnt even a hot take, the majority of professional mathematicians i interact w regularly think bourbaki either sucks or only matters for history (note they're mostly analysts and geometric analysts)
@TedShifrin i dont know how true this story is but one of my profs in my first year said that a big motivation for the bourbakis was that formalizing the shit out of math made it less amenable to applications in war
given when bourbaki was a thing that's at least plausible
Hmmm, is it true that if both the closure and interior of a space are locally compact, then the space itself is locally compact? I think this is probably not true, because otherwise every subset of a locally compact space would be locally compact.
so i guess the conclusion is that maybe (?) the sociology of bourbaki is interesting, but the math is definitely more trouble than it's worth, just read better books that dont kill all the intuition
A: prove that any non-trivial tree actually has at least two vertices of degree one. B: prove that every simple graph with at least two vertices has two vertices of the same degree
Does proving B first immediately make A trivial in the fact that we could classify a non-trivial tree as a simple graph, and by substitution, A is true?
I've been doing too much aviation and too little math lately. :D
I saw the phrase "dihedral group" and thought: yeah, I know what that is. That's a group which I can roll to the right by stomping on the right rudder pedal (and likewise for the left).
If stomping on the right rudder pedal makes the group bank to the left, then it's an anhedral group.
@LeakyNun @Holo: Here is a nice combinatorial problem (that is slightly related to logic): You start with some finite number of stones distributed among some of the vertices of an infinite connected (simple) graph. At each step you must choose some vertex with at least as many stones as neighbours, and move one stone from that vertex to each neighbour. If that is not possible then the game ends. Does the game always end?
Suppose $f:(a,b)\to\mathbb R$, and $f$ differentiable at $(a,b)$. Let $x\in(a,b)$, and $f'$ discontinuous at $x$. Then, $f'(t)$ can't tend to $\pm\infty$ as $t\to x$, right?
@Fargle If $f'(x)$ tends to $\pm$ infinity then $|f'(x)|>|f'(0)|$ for all $x$ with $0<|x|\delta$ for some $\delta$. But this violates intermediate value property.
Hang on, I had a misapprehension here. I was thinking, "Well duh, if f'(t) tends to infinity, then you didn't have differentiability at x in the first place," but that's not the case---say f'(x) just happened to be 3. Oops.
Why the reference to f'(0)? We don't have that (a,b) contains 0 necessarily
@Fargle sorry for that. Rewriting with new info: If $f'$ tends to $\pm$ infinity at $x$ then $|f'(t)|>|f'(x)|$ for $0<|t|<\delta$ for some delta, which violates Darboux Theorem
Oh, I think, to apply that theorem correctly, I should write something like $|f'(t)|>1+|f'(x)|$, and then conclude, right?
@TobiasKildetoft I was thinking of something like $\sqrt[3]{x}$, which has a derivative tending to infinity. But in that case you actually fail to have differentiability at $0$
@MartinSleziak, So, can I conclude from this discussion that, if $f:(a,b)\to\Bbb R$ differentiable at every point in $(a,b)$, and $f'$ discontinuous at $x\in(a,b)$, then $f'$ oscillates from both sides of $x$?
I was supposed to write $\sup_{t\in (x,u)} f'(t) > \inf_{t\in (x,u)} f'(t)$ for every $u>x$?
But in short, I am asking what you mean by saying that $f'$ oscillates.
Of course, it is well-known that a function is continuous at a point iff oscillation at that point is zero. Which is why I mentioned similar expression to the definition of oscillation, with the change that I looked only on one side.
Certainly, if I want oscillation $\omega_f(x)$ to be positive, the positive value, this difference has to be positive on the right or on the left (or both).
This is definition of discontinuity by oscillation I have. So, by Darboux Theorem, I can't have neither jump discontinuity nor removable discontinuity for derivative. And, from discussion before, I can't have $f'$ tending to +- infinity at discontinuity. So, at discontinuity, $f'$ oscillates, right?
@MartinSleziak Wow! you are right :) That is superb text for self study.
If you look at first example - the one with $f(x)=x^2\sin\frac1x$, you can see that you get arbitrarily large positive and negative values of $f'(x)$ close to zero.
So if we use definitions from that book, in this case $f'(x)$ has infinite discontinuity at zero.
You can see this also from the graph of $f'(x)$, which is included in that answer.
I am not sure why Desmos fails so badly on this function, but at the first glance you can see that the red curve is not the graph of this function. Clearly you have $-x^2\le x^2\sin\frac1x \le x^2$ and the graph drawn there is not between these two parabolas.
I see that you do not believe the graphs from Marc McClure's answer. :-)
I'll probably go and get something to eat and then I should get to the stuff I should actually be doing. Still, this was interesting enough that I bookmarked it and also posted in the Calculus and analysis chatroom. Maybe somebody will add something more to this if they notice the discussion there. Thanks for the discussion, @Silent
"Let $f$ be any function from $(-1,1)$ to $\Bbb R$ such that it is differentiable on $(-1,1)$ and, $\lim\limits_{x\to0}f'(x)$ does not exist. We know that $f'(x)$ does not tend to infinity as $x\to0$, since if it did, we would have $|f'(x)|>1+|f'(0)|$ for all sufficiently small nonzero $x$, and this contradicts the intermediate value property of derivatives. Hence there is a sequence $x_n\to0$, $x_n\ne0,$ such that $\lim\limits_{n\to\infty}f'(x_n)=L\ne f'(0)$."
I can't see why there is such an $x_n$ giving $L\ne f'(0)$.
(The intermediate value property of derivatives cited there is Darboux Theorem)
(I have quoted above para from baby Rudin sol manual, for exer 5.19)
Anyway, $T=l[y_1,\ldots,y_n]$, $I$ is a radical ideal and we're assuming $y_1$ is not a zerodivisor of $T/I$ (in fact in the second arrow there is an $y_1$ above, but I don't know how to write it)
Can someone help me understanding what $\frac{T}{I}(-1) $ mean?
Hey, is there a fast way to calculate the residual for a least squares problem? e.g. min ||y-Xb||? I don't particularly care about the coefficients, so I was wondering if there was a better way than just calculating the pseudoinverse?
Suppose i have a diffeomorphism f from R3 to R3 kai M a surface . Why D(fog)=Df*Dg where g is a chart of M has rank 2 ; i know Dg has rank 2 but why the composition also;
I want to prove that f(M) is a surface with fog being the chart but i need the rank of matrix derivative 2 . I know its alot easier to prove that transition maps are smooth but i want to use the definition of a surface
@Secret a 6-simplex is by definition a certain six-dimensional polytope. since what's there is a 3D polytope, that can't be a 6-simplex
(You can view it as the image of a 6-simplex in 3D space with respect to some linear transformation. But that's not anything deep: Every convex hull of n points in m<n dimensional space is obtained from an n-simplex in such a way.)
Yo. In a group, if the inverse of $a$ is denoted $a'$, the commutator $[a,b]$ is often defined as $aba'b'$.
Is there a concise notation for the "double commutator" $ababa'b'a'b'$?
It seems like there ought to be a way to write that concisely using the ordinary commutator notation, but there isn't obviously a more concise way to write it than $ab[a,b]a'b'$.
How I can check the validity of the following argument: (1) If I try and I have talent then I can be a Mathematician. (2) If I become a Mathematician, then I will be happy. ------------------------------------------------------------------- Therefore, if I will not be happy, then I did not try hard or I do not have talent.
my basic stance is: There's a lot of interesting math in the realm of non-self-adjoint spectral problems, and some of it even has applications in physics. But it really doesn't tell you much about quantum theory as such.
As a slogan, "PT-symmetric QM" is fine. As a philosophy, I think it doesn't count for much.
am interested in the "contrarian" aspects of PT symmetry but its rather new to me. wonder if it can be connected with some kind of interpretation issues. suspect the answer is yes but havent seen it yet.
@taritgoswami absolutely! was just hacking on it right now & then tried to start a conversation in chat somewhere :)
Suppose $T \in L(P_3(R), P_2(R))$ is the differentiation map defined by $Tp = p'$. Find a basis of $P_3(R)$ and a basis of $P_2(R)$ such that the matrix of T with respect to these bases is
$ \left( \begin{array}{ccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 \end{array} \right)$
I said th...
