If I have $u_x + u_y = g(y - x)$, the homogeneous solution is $u = f(y - x)$. My understanding is that a particular solution can either be $x\int g(w)dw$ or $y\int g(w)dw$, where $w = y - x$, but when I plug in the guess I'm getting $\int g(w)dw = g(w)$. Should this type of guess be working?
rover, one reason people prove things is because they don't know in advance or via some other route that it is true. if you do, i can see many instances where you might not care about a proof. but often you don't.
and sometimes (not always) a proof can shed light on "why" something is true.
the starting assumptions in many math classes seem 'strange' because they are often designed to be small, if not as small as possible, so you don't have to take large amounts of things for granted. in many applications it is more common to take large amounts of things for granted.
any recommendations i might have would depend on the subject matter you are interested in exploring. i personally think it's very hard to give meaningful advice on proof-writing without remaining specific to one topic, although there are books that try.
start with 'abstract algebra' and not 'analysis' is maybe a good first piece of advice. whether or not it's conceptually simpler, most people find it an easier start.
@leslietownes for now, it's sequences, convergence and divergence. I understand them, but writing it in way of writing proof..seems like writing in other language
mm, i think sequences and series is probably the first time i saw anything resembling proof-like argument, too. it's an adjustment. there you usually eventually have access to a library of 'tests' you can take for granted and the trick is applying them.
tests involving inequalities tend to be harder for beginning students to apply than the tests that don't.
the first proof i remember was that root 2 was irrational.
@Rover well, the idea is that there are some starting facts (axioms) and some rules of inference. the axioms are presumed true and the rules let us determine some 'new' truths and we can keep repeating.
the reality is that we skip most of the stuff and concentrate on the bits we are not sure about.
sometimes proofs are a bit like lock combinations, if you know the combo then it is easy, otherwise you can spent a lot of time scratching your head trying to figure stuff out.
So, a lot of mathematicians nowadays will name things by simply taking old names and adding an intensifier to them. Like, I would not be surprised to find someone had written some paper on something called "ultra-hyperbolic geometry" or "ideals of very strong finite type."
I'm wondering if this is a modern trend, or if people "back in the day" would name things in the same way, but we just don't recognize it as such because it's using older language than we don't immediately understand
interesting thought. my guess would be that it's at least somewhat modern, because if you go back too far, there isn't anything resembling sufficient standardization of core terminology for qualifiers or intensifiers to make sense.
there sort of still isn't, in some absolute sense. but mass publishing, popularity of key books, etc. helped a lot
even in the 1800s people would just say 'function' and mean different things by it, for example. continuity vs. uniform continuity took a while to be figured out, let alone nailed down in those words. maybe by 1900 there was hope of some standardization in some areas.
throwing a guess at the dartboard, i'll declare that mathematicians started doing this in earnest after WW2.
or maybe 1930. WW2 seems a little late for some stuff.
this is something you see a lot in obscure journals (usually with some pay-to-play element). introducing the theory of adjective X, where X is some known thing and adjective is your special sauce. you weaken or strengthen a hypothesis of X and run through what standard textbook stuff still goes through (perhaps with qualifications) and what doesn't. no applications except invented examples for this purpose.
there are real reasons people do that, too, but that's what it made me think of.
good examples of adjective are 'fuzzy' (more popular in 70s-80s) and 'quantum' (works great right now)
i have read and understood the proof of Fubinis theorem multiple times now (about multidimensional integration) what i seem to fail to deveolop is an intuitive understanding, of how the functions defined in the proof connect with the multidimensional integration.. is there some good illustration of this or intutive way of understanding how these functions defined in the proof of fubinis theorem connect to the subject?
anyone have a counterexample for this when $f$ does not have finite order ? let $f$ be analytic in the whole complex plane, such that $f f' f''$ never vanishes. Then $f(z) = \exp(az+b)$ for some constants $a,b$
(its true when $f$ has finite order by the hadamard factorization theorem)
basically it means $f$ cannot grow too quickly at $\infty$, precisely $f$ has finite order if there is some number $a > 0 $ such that $|f(z)| \leq \exp(a|z|)$ for all $z$ sufficiently large
if $f$ has finite order and is entire, there is a result called hadamard factorization theorem that will tell us it has a specific form, and if we also know $f$ has no zeros, the form is exactly $\exp(p(z))$ where $p$ is a polynomial, then knowing in advance that $f'$ has nozeros will tell us that $p$ is necessarily linear
but in the absence of this additional information (namely that $f$ has finite order), I am unsure if this result still holds
what strikes me that this could still be true is that using the hadamard result, i dont think we need to know $f''$ never vanishes, it seems like all we needed is that $f f'$ never vanishes
Hey guys. Can someone tell me how we got from that integral to the lower integral with the given substution? My professor just wrote it and down and said (i dont have time to show you how you do this) i am guessing it is about Polar coordinates. ( i know this from physics) but the problem is , we never learned them in math, so i am guessing maybe they used something else?
@copper.hat im trying to prove that my function is also harmonic by proving it has the mean value property on every ball inside $B_1$ .Now when my ball is entirely on the upper ball or the lower ball obviously satifies the mean value property since $u$ does. Now if my ball is supposingly 3/4 on the upper ball and 1/4 on the lower some terms are killed because $v$ has opposite signs on lower and upper part. But still im left with some terms.
@copper.hat maybe i could just show is satisfies laplaces equation?
@Jam why can't you just use that your thing is harmonic if and only if it satisfies the mean value property locally everywhere, this is clear in the upper and lower (open) half balls, then on the middle lower dimensional ball, you can show that the surface integral evaluates to zero as long as you choose a small enough ball centered at a point in the lower dimensional center ball
because your surface integral is just the integral of $u$ over the upper hemisphere and the integral over the lower hemisphere is the same as the integral of $-u$ over the upper hemisphere by construction
@MadSpaces this is a famous integral. $\int_{[-R, R] \times [-R, R]} e^{-x^2} e^{-y^2} dy dx = \int_{-R}^R \int_{-R}^R e^{-x^2} e^{-y^2} dy dx = \int_{-R}^R e^{-x^2} (\int_{-R}^R e^{-y^2} dy) dx = (\int_{-R}^R e^{-y^2} dy) (\int_{-R}^R e^{-x^2} dx) = (\int_{-R}^R e^{-t^2} dt)^2$ where the interchange of the integrals is permissible due to Fubini
Let V be an inner product space then an operator $T\in L(V)$ is said to be a positive operator if $T$ is self adjoint and for all $v\in V$ the following holds: $\langle Tv, v\rangle \ge 0$. I think that the definition remains unchanged if I remove the condition that "T is self- adjoint". And I think so because if $\langle Tv,v\rangle \ge 0$ for all $v\in V$ then $\langle Tv, v\rangle= \langle v, Tv\rangle$ for all $v$. So by definition of adjoint of $T$, it follows that T is self adjoint.
the conclusion is correct. i'm not sure about your proof. all you would get from the definition of the adjoint is that if <Tw,v> = <w, Tv> held for all w, then T*v = Tv. which isn't what you have.
you'll could snake through a polarization identity or something else that relates a quadratic form to its 'diagonal' to fill in that gap.
i forgot that there is such a thing as a real vector space. :D
for a real vector space something like $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ is a counterexample but if you toss in complex scalars this operator ceases to satisfy the hypothesis.
Ah, I understood. I was concluding only on the basis of $\langle Tv, v\rangle =\langle v, Tv\rangle$ completely disregarding $\langle Tw, v\rangle=\langle w, T^*v\rangle$.
wait... I think my proof is correct. Instead of $\langle ,\rangle$ , I'll use dot .
realness is enough to deduce self adjointness in the complex case, but i do agree that the current argument does not explain why. a polarization formula would complete the picture.
koro as far as i know physicists are the main people who are likely to exclude 'self-adjoint' from the explicit definition of positivity (and also to use positivity of the form). there are twenty zillion characterizations of positivity, some of them quite baroque. it's just simpler to include it in the definition.
koro, another place where this almost comes up is in the notion of ordering on operators. one often writes A <= B if A and B are self adjoint and B-A is positive. it is possible for B-A to be positive even if A and B aren't self-adjoint (e.g. A = P + iQ and B = R + iQ where P, Q, R are self adjoint and Q is nonzero) but people tend not to write <= in that case because <= is just not very interesting on such operators.
lots of nooks and crannies in this area where the definitions seem to build in more hypotheses than necessary, and in some cases this is formally even the case, but it is nevertheless a standard thing to do.
you can tell when an author is in the thrall of some other field of mathematics when they state and prove that a long list of minimal sets of hypotheses are equivalent, and then say "an operator satisfying any of the conditions (a)-(z) is said to be positive." save that kind of goofy definition for some other kind of book.
@Jam you showed it was $C^2$ and Laplace's equation follows. There is no need to do more. In fact, since it matches $u$ on an open set you can say more.
I met a question today: find cardinality of the set of continuous functions $f:[0,1]\to \mathbb R$ such that $\int_0^1 f(x)(1-f(x))\,dx=\frac 14$.
I think that only one such function is possible and that function is $f:x\mapsto \frac 12$ for all $x\in [0,1]$.
Proof: I claim that $f(x)(1-f(x))\le \frac 14$ for every $x$. This is so because maximum value of $t\mapsto t-t^2$ is $\frac 14$. It follows that $\int_0^1f(x)(1-f(x))\, dx\le \frac 14$.
Since integrand is continuous, the equality can hold iff $f(x)(1-f(x))=\frac 14$ for every $x\in [0,1]$. This gives $f(x)=\frac 12$ for every $x\in [0,1]$. So the cardinality of the set is 1.
allie: this may involve dipping into the specifics of how you've introduced the cross product but i'll try to speak to some ideas. because v and w are linearly independent, v x w is a nonzero vector, geometrically, it is a normal to the plane spanned by v and w. if u is perpendicular to that normal, then u is in the plane spanned by v and w.
i have the intuition for why it may be true but i dont know how to prove it
i understand that $v \times w$ is gonna be perpendiculr to the plane spanned by $v$ and $w$ and so $u$ and $v \times w$ cant be perpendicular to eachother since then $\mathbb{R}^3$ would have a dimension of 4 which is obv wrong
so can you convince yourself that u . (v x w) is the determinant with u and then the inputs in it? it looked like you wrote a block matrix of this type (almost?) up above. that's one way to do this.
thanks a lot @copper. Actually, I was trying it in my head for a long time. But everytime I thought I'd solved it -I realized a mistake in the train of thought -For example: my subconscious mind was plotting $t(1-t^2)$ as concave upward. After fixing it, I decided to share the proof.
allie: yeah. when you do the dot product, you just replace i, j, and k with u_1, u_2, and u_3. which symbolically is the same thing as if you'd determinanted the matrix with u in it instead of the ijk thing.
well the "maybe" is what we're getting at with this 3x3 matrix with u, v, w. if its determinant is zero, that tells you something about its rows/columns.
$c_1w.(v\times w)+c_2v.(v\times w)+c_3 ||v\times w||^2=0$ and the first two terms on LHS are zero, whence one gets $c_3=0$ (note that $||v\times w||^2=(v\times w).(v\times w)\ne 0$ because if it is zero then v is parallel to w hence u,v and w can't be LI).
Now, one last step is: since $\mathbb R^3$ has dimension $3$ hence every vector $z\in \mathbb R^3$ can be written as linear combination of $v,w$ and $v\times w$.
What's the difference between strong duality and duality in convex problems? Is it that the first one guarantees the duality for concave functions too?
In particular, there exist some scalars p,q and r such that $u=pw+qv+r(v\times w)$. If $u.(v\times w)=0$ then r=0. This implies that $u=pw+qv$ which is a contradiction as $w,v$ and $u$ are LI. So $u.(v\times w)\ne 0$. QED.
@Koro @leslie I don't know if your positivity discussion got resolved, but there's no way to get from what you had to what you need (polarization does not work). Moreover, you have to be suspicious since it fails over $\Bbb R$ but you can certainly consider matrices on $\Bbb C^n$ with all real entries.
I am trying to prove that $K(x,y) = \cos(x-y)$ is a PSD kernel on $[0,2 \pi] \times [0,2 \pi]$. The hint I am given is to consider $\phi : [0,2 \pi] \to \Bbb{R}^2$ given by $\phi(x) = (\cos (x), \sin (x))$. But I'm not sure how to proceed from this hint.
this might not be the right identity (looks antilinear in the first coordinate :(() but i had in mind something similar to math.stackexchange.com/questions/561636/…
Hmm ... While I was at PT I certainly convinced myself (working in my head, no paper) that the polarization cannot work over $\Bbb R$. I am still suspicious what happens to Daniel's argument if you start with a real non-symmetric positive matrix.
Oh, that came up in the comments.
This might be one of the only linear algebra results I know of that works $/\Bbb C$ but fails $/\Bbb R$.
it's just wrapped in something that only physicists care about. in the guise of a statement about invariant subspaces and eigenvalues it is less surprising.
I think I once figured out that positive issue in the comment, many years ago when this discussion came up. I only define positivity in the self-adjoint setting.
every math book i have ever seen includes it in the definition. i think for some physics folks the quadratic form comes first and so they come at it differently.
professor Ted, I realized earlier during discussion with leslie that the proof (Over C, an operator is self adjoint iff $\langle Tv, v\rangle$ is real for all v) that I'm aware of (from Axler's) works when we are in field of complex numbers.
Comment to @allie: In my book, we only find out in Chapter 7 that linear independence of $n$ vectors in $\Bbb R^n$ is determined by nonzero determinant. In Chapter 1 that is established for $2$ vectors in $\Bbb R^2$. The right argument is what was given above. If $u$ is orthogonal to $v\times w$, then $u$ lies in the plane spanned by $v$ and $w$.
@Koro I guess that's a good tricky exercise, but as I've said, to me positivity is only relevant for self-adjoint operators in the first place. :P
this is one reason why wikipedia is a disaster for learning about this stuff. tons of related notions, with definitions that sometimes are and sometimes are not equivalent, scattered across pages written and edited by different people at different times.
Lots more than that, but it's a question you — for example — should contemplate. It comes up in complex geometry because the real part of a hermitian metric gives the associated Riemannian metric and the imaginary part is the associated Kähler form (i.e., is alternating).
He has lower-level books, too. The main virtue of Strang is the good treatment of interesting applications. It's not the place to learn proof-based linear algebra.
koro, one thing you sometimes see in complex * -algebras is the definition that T is positive if there's some B in the algebra with T = B* B. this is equivalent to the usual definition if the algebra is complex matrices, or bounded operators on a complex hilbert space. but if the elements your algebra isn't given as necessarily 'acting' on anything (let alone a hilbert space) it's a bit of work to relate this condition to other forms of positivity.
for example i don't think it's immediate that a sum of positive elements is positive, under this definition.
some books in this area basically apologize for introducing the complex conjugate in the hermitian inner product, saying stuff like 'just trust us, it makes <x,x> nonnegative' or 'we will come back to this later' or 'physics demands this'
the best approach is just to push those symbols onto the page, and then into readers' heads
do we need the optimization problem to be convex for KKT conditions to hold? I'm trying to thing of a non-convex problem where KKT are still guaranteeing optimality, but I guess it's the h(x) = 0 condition that would be violated then?
You can understand my comments, @Koro, if you wish to. Look at the real part of the hermitian inner product of two vectors $z$ and $w\in\Bbb C^n$. See that it is the real inner product if you view those as vectors in $\Bbb R^{2n}$. So, in particular, hermitian orthogonal implies real orthogonal. But there's more, obviously. What does the imaginary part tell you?
Oh, actually, maybe Balarka's equalities aren't orthogonal to my comments. I'd need to think, and that takes work.
maybe a little bit more precise, maybe I should've said: "after image" of a tangent vector parallel transported along a curve for a short period of time.
that's one heck of an abbreviation. L doing the work of linear, and erp doing the work of intERPolation. i guess nobody wanted to write lint although there the 'in' could apply to both words.
I am finding the sine of the acute angle between the vectors a = 2i + j + 2k, b = -3j + 4k using the cross product. The textbook says to substitute |nhat| = 1 into the formula a x b = |a||b|sin(theta)nhat. Firstly, the formula uses nhat which is a vector, not |nhat|. Secondly, why can you substite |ncap| = 1?
Realization: the quadratic Bézier curve with control points $a$, $b$, and $c$ is the same as the cubic Bézier curve with control points $a$, $(a+2b)/3$, $(2b+c)/3$, and $c$
