User zen harper - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:53:05Z http://mathoverflow.net/feeds/user/6651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98451/understanding-the-inverse-laplace-transform-of-a-function-with-essential-singular/98562#98562 Answer by Zen Harper for Understanding the inverse Laplace transform of a function with essential singularities Zen Harper 2012-06-01T10:28:53Z 2012-06-01T10:38:21Z <p>After taking a quick look at the paper, I agree with Robert Israel. </p> <p>Almost no detailed justification is given (<em>perhaps not surprising for a Physics journal...</em>), and it seems to be not so easy to justify.</p> <p>Take the special case $N=1$ for simplicity; so up to constants, we consider (note the use of (A3) in the paper, this constant is subtracted from the original $\tilde{I}$)</p> <p>$$\tilde{I}(s) = \frac{(1-\exp(-Ks/(s+r))}{s/(s+r)} - (1-\exp(-K))$$</p> <p>for $K$, $r$ constants. Now consider what happens as $|s| \to \infty$, then we get [assuming, dangerously, that my calculations are error free]</p> <p>$$\tilde{I}(s) \sim -Kr e^{-K}/(s+r) \sim A/s$$</p> <p>as $|s| \to \infty$, for some constant $A$. The contour is originally a vertical line, but $1/s$ is not in $L^1$ so the integral is not a standard Lebesgue integral (i.e., is not absolutely convergent). However, it is is $L^2$, i.e. is square integrable, so you can use the $L^2$ theory of the Fourier transform to give a meaning.</p> <p>Now, if you deform the original vertical line contour over { Re(s) = c } by Cauchy's Theorem (push part of it to the left), you get the integral of $\tilde{I}(s) e^{st}$ over the circle $C$ plus an error term, being the sum of integrals over the contours</p> <p>C1 = { $L + iy : |y| &lt; R$ }, </p> <p>C2 = { $x \pm i R : L &lt; x &lt; c$ },</p> <p>C3 = { $c + iy : |y|>R$ }</p> <p>where $c$ is fixed; now let $L \to -\infty$ and $R \to +\infty$ appropriately to get the error term going to zero.</p> <p>[More detail: as $R \to \infty$ the integral over C3 goes to 0 uniformly in $t$, by the $L^2$ properties of the Fourier transform; also the integral over C2 goes to zero because $|1/s| &lt; 1/R$ on C2.</p> <p>The integral over C1 goes to zero as $L \to -\infty$ because $| e^{st} / s | &lt; e^{Lt}$ on C1, which clearly $\to 0$ rapidly, at least for $t>0$. ]</p> <p>There are still 2 things to be justified in this special case: (i) what about $t&lt;0$, and (ii) how do we know the analytic function $\tilde{I}(s)$ really can be represented as the Laplace transform of something?</p> <p>I think (ii) follows from a general result in my last paper (about Laplace transform representation theorems, which is in Documenta Mathematica 2010, or on my website); but (i) I am not sure about.</p> <p>[<em>Strangely enough, the contours C1, C2, C3 are exactly the ones I used in my paper, although I got them from an earlier paper by C.Batty and M.D.Blake; and I suspect the original use of these contours dates back at least 50 years, since similar stuff has been used in Tauberian theory and analytic number theory for a long time</em>].</p> <p>So, even in this special case with just one singularity, there are a lot of extra details to fill in if you want to make it properly rigorous. Since the original paper is from a Physics journal, I would guess it's not the best place to look for rigour(!) Approach with caution...</p> http://mathoverflow.net/questions/56257/parametrisations-for-null-temperature-functions-nonuniqueness-of-solutions-to-th Parametrisations for Null Temperature Functions: nonuniqueness of solutions to the Heat Equation Zen Harper 2011-02-22T08:42:36Z 2011-12-01T00:38:06Z <p><strong>Disclaimer</strong> <em>I expect this is a highly open problem, but maybe I'm wrong and someone has come up with some answers besides those given here. In any case, all information appreciated, thanks!</em></p> <p><strong>Definition</strong> A <em>null temperature function</em> is a continuous function $u = u(x,t) : \mathbb{R} \times [0, \infty) \to \mathbb{R}$ such that the <em>heat equation</em> is satisfied on the interior, i.e. $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} \quad \text{ for } t>0,$ but $u(x, 0) \equiv 0$ for all $x \in \mathbb{R}$.</p> <p><em><strong>EDIT:</em></strong> if you wish, assume all partial derivatives (of all orders) to exist and be continuous also [in the interior]; I expect this doesn't actually make much difference, by elliptic regularity arguments.</p> <p><strong>Theorem</strong> There exists a null temperature function satisfying $|u(x,t)| &lt; \exp(A/t)$ with $A>0$, such that $u(x,t) \not\equiv 0$ for some $t>0$.</p> <p><strong>Theorem</strong> Let $u$ be a null temperature function satisfying $|u(x,t)| \leq A \exp(B t^{-\delta})$, for some $A,B>0$ and $\delta&lt;1$. Then $u \equiv 0$.</p> <p><em>[I should mention that these are not my theorems! References:</em></p> <p><em>S.-Y. Chung, D. Kim. An example of nonuniqueness of the Cauchy problem for the heat equation. Comm. Partial Differential Equations 19 (1994), no. 7-8, 1257–1261. MR1284810 (95c:35114)</em></p> <p><em>S.-Y. Chung. Uniqueness in the Cauchy problem for the heat equation. Proc. Edinburgh Math. Soc. (2) 42 (1999), no. 3, 455–468. MR1721765 (2000h:35060)</em></p> <p><em>Also discussed briefly, with connections to the uniqueness problem for the Laplace transform, in my paper "Laplace transform representations and Paley–Wiener theorems for functions on vertical strips"]</em></p> <p><em><strong>Vague Questions</em></strong> Besides the results above, what is known about the class of null temperature functions? Clearly it is a vector space; can it be given a "natural" Banach space norm? Can we represent it (or nice subspaces of it) in any nice way? What kind of growth rates are possible?</p> <p><strong>EDIT</strong> <em><strong>Precise question - maximal growth rates</em></strong> Is there some universal function $\varphi : (0,1) \to \mathbb{R}$ with the following property? </p> <p>For every non-trivial Null Temperature Function $u$ such that $M(t) = \sup_x |u(x,t)| &lt; \infty$ for each $t>0$, there is some $C &lt; \infty$ such that $M(t) \leq C \varphi(t)$ for all $t \in (0,1)$.</p> http://mathoverflow.net/questions/67224/why-is-frac-sqrt63229-sqrt145-approx-pi Why is $\frac{\sqrt{6}}{32}(29 + \sqrt{145}) \approx \pi$ ? Zen Harper 2011-06-08T08:41:54Z 2011-06-11T21:24:58Z <p><em><strong>Apologies in advance</em></strong> if this is a stupid question; also, <strong><em>disclaimer: this is purely for fun</em></strong>; but:</p> <p>Why is $\frac{\sqrt{6}}{32}(29 + \sqrt{145})$ such a good approximation to $\pi$?</p> <p>(<em>Correct to 8 decimal places if my calculator is to be believed</em>).</p> <p>Is it just a numerical coincidence, or is there actually some deep explanation (like the famous almost-integer $e^{\pi\sqrt{163}}$ related to Heegner numbers)?</p> <p>[Not my observation, though; this is taken directly from a comment by the user "unknown (yahoo)" in the closed question</p> <p><a href="http://mathoverflow.net/questions/67161" rel="nofollow">http://mathoverflow.net/questions/67161</a></p> <p>...who noticed that $\frac{9}{8\pi} + \frac{8\pi}{29} \approx \sqrt{3/2}$, which is equivalent to the approximate equation above; although he/she didn't give any further explanation].</p> <p>My knowledge of number theory is extremely limited, but it looks like standard methods like continued fractions etc. won't work here - or will they?</p> <p><strong>* EDIT: after some more numerical searches: *</strong></p> <p>$(\sqrt{a} + \sqrt{b})/c$ with integer $a,b,c$ can approximate a bit better for certain $c$ than for others; e.g. for $c=32$ as above, the error of the best approximation is (very) roughly $1/100$ of the error for $c= 23,26,29,30$ and $1/10$ of the error for $27,28,33,34$, etc.</p> <p>But all values of $c$ seem to be not too bad; the worst values have errors still within a factor of roughly $100$ of the best values. Results with $e$ and random numbers instead of $\pi$ seem vaguely similar.</p> <p>So, this approximation is maybe not as striking as I first thought; but anyway, maybe there is still something deeper lurking behind it. Maybe, also, it's nothing to do with $\pi$.</p> http://mathoverflow.net/questions/38245/what-is-the-easiest-randomized-algorithm-to-motivate-to-the-layperson/63872#63872 Answer by Zen Harper for What is the easiest randomized algorithm to motivate to the layperson? Zen Harper 2011-05-04T03:50:05Z 2011-05-06T00:46:21Z <p>[<strong>EDIT: disclaimer! I probably shouldn't have posted this answer, because it's nowhere near my area of expertise; so beware, it may contain nonsense!</strong>]</p> <p>[<strong><em>EDIT: sorry; I saw this nice problem in a book, but it seems a nice deterministic algorithm in $O(n (\log n)^4)$ time was found later, so this answer is inapplicable; reference:</em></strong></p> <p><em>Matching nuts and bolts</em> by Noga Alon, Manuel Blum, Amos Fiat, Sampath Kannan, Moni Naor, Rafail Ostrovsky;</p> <p>citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.103 ]</p> <p>The "nuts and bolts" problem, which I saw just a few days ago in the Algorithm Design Manual [precise reference to follow when I get time to edit]:</p> <p>You are given $n$ pairs of nuts and bolts, all very slightly different sizes (so, each nut fits exactly one bolt); but unfortunately they have all been disassembled, and you have to fit them together.</p> <p>You cannot see the difference in size by eye, so the only way to see if a nut fits a bolt is to try it. If a nut is too large/small for a bolt, you will discover this; but note that <em>you can't compare two nuts, or two bolts, with each other directly</em>.</p> <p>Abstractly: we have 2 real unknown sequences $(x_1, x_2, \ldots, x_n)$ and $(y_1, y_2, \ldots, y_n)$.</p> <p>The $x_i$ are all distinct, and $(y_j)$ is an unknown rearrangement of $(x_i)$. We want to match them up, but the ONLY measurement we can make is to pick $i,j$ and compare $x_i$ with $y_j$.</p> <p>The obvious algorithm of trying all of them one-by-one takes $O(n^2)$ time.</p> <p>However, a variant of <em>randomised</em> quicksort takes $O(n \log n)$ time on average. The randomness just comes from picking each nut/bolt at random.</p> <p>Details: pick a <em>nut</em>, use that as the pivot element for the <em>bolts</em>; then use the matching <em>bolt</em> as the pivot element for the <em>nuts</em>, etc. - like quicksort, but going back-and-forth between $x$ and $y$ elements.</p> http://mathoverflow.net/questions/60457/elementaryshortuseful/60670#60670 Answer by Zen Harper for Elementary+Short+Useful Zen Harper 2011-04-05T10:56:43Z 2011-04-06T08:02:27Z <p>Uniform convergence of the averages of the partial sums of the Fourier series, for any continuous function $f$ on $[0, 2 \pi]$ with $f(0)=f(2\pi)$:</p> <p>$$\sigma_N(f, \theta) = \sum_{n = -N}^N \left(1-\frac{|n|}{N+1} \right) \widehat{f}(n)e^{in \theta} \to f(\theta)$$</p> <p>And the Weierstrauss Polynomial Approximation Theorem: the polynomials are uniformly dense in $C[a,b]$. This is a corollary of the Fourier series result, or it can be proved similarly. Finally, if time permits, the Stone-Weierstrauss Theorem.</p> <p>Of course, it would be nice to talk about approximations to the Dirac Delta, convolutions, fundamental solutions to PDEs, e.g. the Heat Equation, etc. etc. but I suppose only a REALLY good class could absorb all this in half an hour...</p> http://mathoverflow.net/questions/32566/careers-advice-for-ph-d-s-without-current-postdocs-or-university-jobs Careers advice for Ph.D.s without current postdocs or university jobs Zen Harper 2010-07-20T00:42:21Z 2011-03-30T03:07:26Z <p>Hi,</p> <p>I'm sure I'm not the only Ph.D. mathematician on MO in serious need of career advice. I'm sure there will be other readers in similar situations, who will find any good advice very helpful. Can anyone suggest anything? <strong><em>Honest, serious</em></strong> answers only please. </p> <p>Note that the <strong>obvious</strong> advice, i.e. <em>do lots of great research, write loads of papers, make friends with lots of professors at conferences and seminars, apply to many jobs, learn loads of new topics, get a brain upgrade</em>, etc. etc. etc. is already known to me and most other people on MO.</p> <hr> <h3>Background motivation</h3> <p>Suppose someone (who shall remain anonymous, but let's call him <em>Dr.H</em> for the sake of argument) is in the following position:</p> <p><em>Dr.H</em> has a Ph.D. in Pure Mathematics from a good English university.</p> <p><em>Dr.H</em>'s Ph.D., whilst perfectly respectable from the mathematician's viewpoint, is not known to be of any use for industrial research or non-university jobs of any kind.</p> <p><em>Dr.H</em> has several years' postdoc/lecturing experience, but only at universities with very low academic reputations, which has now ended.</p> <p><em>Dr.H</em> has several published papers in good journals; but unfortunately, less than other people of his age in his area. He can do good work, but too slowly.</p> <p><em>Dr.H</em> currently has a non-university, non-research job teaching in a school, and cannot easily attend conferences, seminars, university libraries, etc. etc., and consequently <em>Dr.H</em> now has even less time for research than before.</p> <p>Most advertised mathematical jobs (e.g. www.math-jobs.com, www.jobs.ac.uk), both academic and non-academic, demand teaching experience or other skills which <em>Dr.H</em> does not possess, and does not know how to acquire.</p> <p><em><strong>SUMMARY:</em></strong> <em>Dr.H</em>'s research record is quite good, but it seems not good enough for <em>Dr.H</em> to get a university job involving research. However, <em>Dr.H</em>'s teaching experience also seems not to be good enough to get a purely teaching university job. So <em>Dr.H</em> appears to be in a very tricky situation!</p> <hr> <h3><em><strong>Question</em></strong>: what should <em>Dr.H</em> do?</h3> <p>Does <em>Dr.H</em> have any reasonable chance of continuing his academic career? If so, how? (<em>Apart from the obvious "apply for more jobs, publish more papers"</em>).</p> <p>Should he apply to advertised university jobs, even though he does not satisfy the requirements? Isn't this simply a waste of time?</p> <p>Or should <em>Dr.H</em> abandon the universities entirely and seek non-university jobs?</p> <p>Does <em>Dr.H</em> have any real advantage over new B.Sc. Mathematics graduates when applying for non-university jobs of relatively low mathematical content? If so, how should he find such jobs, what exactly are these advantages, and how should he make full use of them?</p> <p><em><strong>Important note:</em></strong> in your answers, please state <strong><em>which country</em></strong> you are referring to, since this can make a big difference!</p> http://mathoverflow.net/questions/57794/undefined-gamma-function-problem/57800#57800 Answer by Zen Harper for Undefined gamma function problem Zen Harper 2011-03-08T09:33:44Z 2011-03-08T09:39:43Z <p>A direct calculation for $d=2$ is also possible and interesting:</p> <p>Let $G(\lambda, \mu) = \int_0^\infty \frac{e^{- \lambda t} - e^{- \mu t}}{t} dt$, for $\lambda, \mu > 0$. There are no problems at $t=0$ because $e^{- \lambda t} - e^{- \mu t} = (\mu - \lambda)t + O(t^2)$ near zero.</p> <p>Clearly $G(\lambda, \mu) = G(\lambda/\mu, 1)$ by a substitution, so we need only calculate $F(\lambda) = G(\lambda, 1) = \int_0^\infty \frac{e^{- \lambda t} - e^{- t}}{t} dt$.</p> <p>Now by differentiation under the integral sign,</p> <p>$$F'(\lambda) = -\int_0^\infty e^{- \lambda t} dt = -1/ \lambda$$</p> <p>Since clearly $F(1) = 0$, we have $F(\lambda) = -\log \lambda$, so (<em>as Anatoly Kochubei correctly says in another answer</em>), the final answer is $G(\gamma, \delta) = -\log(\gamma/\delta) = \log \delta - \log \gamma$.</p> <p>(<em>Note that these calculations are all easy to justify rigorously, by simple estimates.</em>)</p> http://mathoverflow.net/questions/56064/fourier-series-application-for-dissertation/56134#56134 Answer by Zen Harper for Fourier Series application for dissertation Zen Harper 2011-02-21T03:13:27Z 2011-02-21T03:13:27Z <p>Fourier series are useful (and sometimes essential) for solving/understanding many problems involving periodic functions on $\mathbb{R}$ or, equivalently, functions $f$ on $[a,b]$ such that $f(a)=f(b)$. I was going to say <em>almost all problems</em>, but that's probably an exaggeration. Of course it helps if the problem is linear, and the properties of the functions you're considering can be easily expressed in terms of the Fourier coefficients -- but even then these restrictions are not always essential. </p> <p>e.g. the Heat Equation on a (physical) ring, where periodicity is assured by the shape of the space; (Willie Wong already mentioned this in the comments).</p> <p>My favourite one: proving the Isoperimetric Inequality, that the circle has the largest area of all piecewise $C^1$ curves with given perimeter;</p> <p>The functional equation for the Riemann Zeta Function $\zeta$: one proof involves the Fourier expansion of the sawtooth function $x - [x]$, which I think I saw in E. C. Titchmarsh's old book <em>The Theory of the Riemann Zeta Function</em> (although I'm sure many other books will give it also).</p> <p>I think it was either Hardy or Littlewood (or maybe both?!) who said that a periodic function should <em>always</em> be expanded as a Fourier series; if you always follow this rule then it'll solve a lot of problems automatically!</p> <p>Although one should be cautious; "<em>if the only tool you have is a hammer, then everything looks like a nail</em>"...</p> http://mathoverflow.net/questions/55715/are-there-uncountably-many-essentially-inequivalent-versions-of-mathematics Are there uncountably many essentially inequivalent versions of Mathematics? Zen Harper 2011-02-17T10:06:43Z 2011-02-17T22:59:54Z <p>Hi everyone,</p> <p><strong>Disclaimer 1:</strong> <em>logic and set theory are a long way from my field, so apologies in advance if I demonstrate extreme ignorance or stupidity, and please correct me if (when?) I write stupid things. But hopefully my basic meaning should be fairly clear to everyone even if I get some details wrong.</em></p> <p><strong>Disclaimer 2:</strong> <em>I admit this question might be slightly subjective. But I feel it's not too subjective, and is fairly natural and interesting to most mathematicians, out of mere curiosity.</em></p> <p><strong>Framework:</strong> Throughout, let's assume that standard ZF set theory is consistent, and take it as our basic mathematical foundation. (<em>I don't necessarily think this is best, but I prefer to pin down the discussion</em>).</p> <p>We all know that Mathematics comes in several distinct flavours: e.g. you can believe or disbelieve the Continuum Hypothesis, and both points of view are (equally?) valid; they are really just matters of opinion. Thus there are at least 2 different versions. Of course we have <em>infinitely many</em> different versions: each number $m=1,2,3,\ldots$ gives a different flavour of Mathematics, given by the axiom $2^{\aleph_0} = \aleph_m$.</p> <p><strong>Subquestion</strong> Does the value of $m$ really matter very much? $2^{\aleph_0} = \aleph_1$ seems a particularly special case; but I find it hard to believe there'd be very much meaningful distinction (<em>in terms of theorems anyone would want to consider</em>) between the axioms $2^{\aleph_0} = \aleph_{103}$ or $2^{\aleph_0} = \aleph_{275}$, for example.</p> <p><strong>If desired</strong>, we could regard these different versions of Mathematics as essentially equivalent (<em>in a rough sense</em>): the axioms all look very similar, given by a single parametrisation. We could also throw in versions with $2^{\aleph_\alpha}$, etc.</p> <p><strong>Alternatively</strong>, we could remove these difficulties completely by not even considering cardinals beyond $\aleph_2$ or $\aleph_3$, say; (or any $\aleph_m$ with finite $m$). </p> <p>It would be really amusing if we could do the following, for then we would have (at least) $2^{\aleph_0}$ different flavours of Mathematics! (<em>Although I suppose there might be technical difficulties with nonconstructive infinite 0,1 strings...!</em>) We'd have an explicit injective function $f$ from $[0,1]$ into the class of all possible versions of Mathematics!</p> <h1>Main question</h1> <p><strong>Can we find</strong> (or prove the existence of) an <em>infinite</em> sequence of axioms $A_1, A_2, A_3, \ldots$, for which <em>every</em> sequence of true/false assignments is consistent? (e.g. the infinite string 1011001110... would mean that $A_j$ is true for $j=1,3,4,7,8,9,\ldots$ and false for $j=2,5,6,10,\ldots$; we want every string to be consistent).</p> <p><strong>If so</strong>, can it be done with $A_1, A_2, \ldots$ all being essentially different kinds of axioms? [<em>maybe it's stupidly optimistic to hope for this</em>]. Can it be done without ever considering $\aleph_k$ for $k>3$, say (or 4, or any fixed finite number)?</p> <p><strong>If not</strong>, what's a reasonable known lower bound $K$ on the number of $A_1, \ldots, A_K$ which are known to exist, so that we have at least $2^K$ essentially different versions of Mathematics?</p> http://mathoverflow.net/questions/53373/the-frechet-derivative-and-lagrange-multipliers-on-banach-spaces/53397#53397 Answer by Zen Harper for The Frechet derivative and Lagrange multipliers on Banach spaces Zen Harper 2011-01-26T20:18:45Z 2011-01-26T20:18:45Z <p><strong>NOTE</strong>: this was a comment, because I thought it wasn't detailed enough for an answer; but Jules (the OP) specifically asked me to post it as an answer. </p> <p><strong>NOTE to Jules</strong>: However, maybe you should wait a few hours or days before accepting any answer, to give others a chance to read it (differing time zones around the world, etc.) [<em>Some MO people seem to get a bit annoyed when people accept answers very quickly</em>].</p> <p>Sorry I can't think of any particular book to recommend for Calculus of Variations; but I think an advanced undergraduate book (maybe just a chapter or two of a "Mathematical Methods" book) might be more useful than a detailed Graduate level book, which would probably be more than you need.</p> <p>It depends what you mean by a solution, and the level of rigour you want. Physicists and engineers solve similar problems non-rigorously all the time, by Calculus of Variations methods (which involve Lagrange multipliers). Euler-Lagrange equations are the keywords to search for. However, they usually don't specify exactly which functions they consider. Banach spaces only arise if your restrictions are explicit: being $C^1$, $C^2$, etc., and you define a complete norm on the class of functions. But for most specific problems, Banach space theory is probably not worth the effort.</p> http://mathoverflow.net/questions/53352/magic-square-in-the-complex-plane-with-equal-integrals-along-every-horizontal-ve/53371#53371 Answer by Zen Harper for magic square in the complex plane with equal integrals along every horizontal, vertical and diagonal Zen Harper 2011-01-26T16:22:36Z 2011-01-26T18:45:11Z <p><strong>CORRECTION</strong>: many thanks to Joel David Hamkins and Willie Wong; yes, we should take $I_1 = I_2 = 0$ (now edited out and removed). For some reason "constant" and "zero" keep getting mixed up together in my head!!</p> <p>You can get lots of examples satisfying the vertical and horizontal integral conditions by choosing $f(x,y) = g(x)h(y)$ where $\int_0^1 g(x) dx = \int_0^1 h(y) dy = 0$. The diagonal integrals being equal to $0$ gives you two equations relating $g,h$, but these still leave very many possibilities for $g,h$.</p> <p>(Note: I wrote $I_1$, $I_2$ before, but I've removed these to make it clearer).</p> <p>Finally, taking arbitrary finite linear combinations of such $f$ (<em>and also infinite linear combinations, if you are careful with convergence</em>) gives you yet more examples.</p> <p>If you want to restrict $f$ to, say, analytic functions, then this won't work - but you didn't say this in your question! (Although I suppose maybe you meant this, since you do say "analytic" magic square!)</p> <p>EDIT: I'm definitely <em>not</em> claiming that <em>every</em> continuous example $f$ can be obtained in this way!! I just want to show that there are many, many possibilities for $f$.</p> http://mathoverflow.net/questions/52032/examples-of-non-metrizable-spaces/52045#52045 Answer by Zen Harper for Examples of non-metrizable spaces Zen Harper 2011-01-14T07:02:30Z 2011-01-14T07:16:38Z <p>I'm not an expert, but I believe the space of Distributions (in any number of variables), (a.k.a. generalised functions, including the Dirac delta, its derivatives, etc.) as used in PDE theory, is a topological vector space, but <em>non-metrisable</em>; even though sequences <em>are</em> sufficient to do everything.</p> <p>However, the subspace of <em>tempered</em> (Schwartz) distributions, as used in Fourier Analysis, <em>is</em> metrisable; it is a Fréchet space.</p> http://mathoverflow.net/questions/51531/theorems-that-are-obvious-but-hard-to-prove/51544#51544 Answer by Zen Harper for Theorems that are 'obvious' but hard to prove Zen Harper 2011-01-09T14:10:42Z 2011-01-09T14:10:42Z <p>Maybe on the boundary of what's allowed, but I would say most basic geometric things like Pythagoras' Theorem, trigonometry with sine/cosine, the area of a circle, etc. etc. etc.; here of course the difficulty is in <em>defining</em> what we mean by length, area, angle, etc. - in which case some of these become axiomatic, but then the difficulty is shifted onto proving that things do work correctly.</p> http://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-function/48149#48149 Answer by Zen Harper for When I can safely assume that a function is a Laplace transform of other function? Zen Harper 2010-12-03T07:53:19Z 2010-12-03T07:53:19Z <p>This kind of question is very interesting, and I too would like to know answers.</p> <p>Sorry to self-publicise; I hope it's not regarded as impolite, but since I have also considered this exact kind of question, it's quickest just to refer to my own paper (<em>and the references I give in there</em>):</p> <p><strong>Laplace Transform Representations and Paley-Wiener Theorems for Functions on Vertical Strips</strong> by Zen Harper. <em>Documenta Math. 15 (2010) 235-254</em>.</p> <p>This is <em>freely available online</em> from the journal:</p> <p><a href="http://www.math.uiuc.edu/documenta/vol-15/vol-15-eng.html" rel="nofollow">http://www.math.uiuc.edu/documenta/vol-15/vol-15-eng.html</a></p> <p>Given an analytic function on a vertical strip, I try to find conditions which guarantee it can be represented by a bilateral Laplace transform (in various senses). I definitely don't claim to have any kind of complete answer, but it's the best I know of (<em>by definition! If I knew any better answers, I would have written them in my paper!</em>)</p> <p>It seems like there are still many open questions about this.</p> http://mathoverflow.net/questions/45802/undergraduate-math-research/45806#45806 Answer by Zen Harper for Undergraduate math research Zen Harper 2010-11-12T09:55:11Z 2010-11-12T09:55:11Z <p>You say:</p> <p>"<em>I am a junior at the moment and taking: One Dimensional Real Analysis, Intro to Numerical Methods, and Abstract Algebra</em>".</p> <p>Based on this information, I think it is a <strong>complete waste of your time</strong> even to consider research seriously at this stage; you need several more years of study as a minimum. Right now, you are still learning the basic language of mathematics. It's similar to, say, a student who wants to begin reading classical German literature, but only knows 100 words -- premature, to say the least. The maths you know right now is probably less than 1% of what you will need. Even after my Ph.D., I feel that my knowledge is very limited in comparison to most good researchers.</p> <p>But do you really mean "research", i.e. new, original, nontrivial and interesting, and publishable in a good quality journal, i.e. one which your professors would publish in?</p> <p>Or do you mean a kind of "investigation" or "project" instead? These are not required or expected to contain anything new or original. This would be highly worthwhile -- but only for your personal interest and satisfaction.</p> <p>The question is, what do you expect to get out of it? If you're at a good university, their lecture courses should already provide you with all you need.</p> <p>Please don't take offense, and apologies if I've formed the wrong impression, but it sounds to me (<em>from your statement "I have begun speaking with professors about their research also"</em>) like you might be the kind of student that irritates professors, always bugging them and asking them questions about their own research, but lacking the knowledge to understand the answers. (<em>But it's not your fault you lack knowledge - that's what you're at university to learn!</em>) As an analogy, imagine a ten-year-old, knowing nothing more than how to add fractions, constantly harrassing you to teach them about calculus; my response (unless I were in a very good mood that day) would be: "<em>go back to school and stop bothering me, for at least another 3 years!</em>" Unless you're an exceptionally good, enthusiastic student, or your professors are far more patient than me, that might be what they're thinking also, but are too polite to tell you.</p> <p>But just my opinion, don't take my word for it; why don't you ask them directly if that's what they're thinking?!</p> http://mathoverflow.net/questions/38049/what-exactly-has-louis-de-branges-proved-about-the-riemann-hypothesis What, exactly, has Louis de Branges proved about the Riemann Hypothesis? Zen Harper 2010-09-08T12:06:42Z 2010-09-12T13:01:50Z <p>Hi,</p> <p>I know this is a dangerous topic which could attract many cranks and nutters, but:</p> <p>According to Wikipedia [<em>and probably his own website, but I have a hard time seeing exactly what he's claiming</em>] Louis de Branges has claimed, numerous times, to have proved the Riemann Hypothesis; but clearly few people believe him. His website is:</p> <p><a href="http://www.math.purdue.edu/~branges/site/Papers" rel="nofollow">http://www.math.purdue.edu/~branges/site/Papers</a></p> <p>but I find his papers difficult to follow. However, whether or not you believe him, his arguments presumably should prove <em>something</em>, even if not the full RH.</p> <p>So, my question is:</p> <p><em><strong>Are there any theorems related to the Riemann Hypothesis and similar problems, arising from his work, which have been fully accepted by the mathematical community and published (or at least submitted)?</em></strong></p> http://mathoverflow.net/questions/35538/the-maximum-of-a-real-trigonometric-polynomial/35826#35826 Answer by Zen Harper for The maximum of a real trigonometric polynomial Zen Harper 2010-08-17T01:49:13Z 2010-08-17T01:49:13Z <p>Even in the special case where $f(x) \geq 0$ for all $x$, there can't be any simple answer involving the coefficients $(a_n)$, $(b_n)$. You're basically asking to estimate the $L^\infty$ norm of a trigonometric polynomial in terms of the Fourier coefficients, and it's well known that this can't be done in any good way (more generally, the relation between the $L^p$ norm and the coefficients is horribly intractable, for any $p \ne 2$).</p> <p><em><strong>EDIT:</em></strong> I suppose it depends what you mean by a "good" way to approximate; this is a bit subjective, but I think "for any reasonable purpose" (any general-purpose programme you would actually run on a computer) no simple <em>theoretical</em> formula exists (which is <em>guaranteed</em> to have good error bounds).</p> <p>However, if you want a <em>numerical</em> scheme to approximate a <em>specific</em> polynomial, that's a totally different question! You need a good numerical analyst (<em>which I am not, sorry</em>).</p> http://mathoverflow.net/questions/35443/coefficients-of-holomorphic-functions-defined-by-borel-probability-measures-on-th/35463#35463 Answer by Zen Harper for Coefficients of holomorphic functions defined by Borel probability measures on the unit disc Zen Harper 2010-08-13T09:52:18Z 2010-08-16T02:33:11Z <p><em><strong>EDIT: this was for a different problem, but it has now been changed; so ignore this!</em></strong></p> <p>Too long for a comment...</p> <p>If the integral is over $D$, then $f'(z)$ is (after changing variable $w$ to $\bar{w}$ in the measure $\mu$) $\frac{d}{dz} \int_D \frac{d\mu(w)}{1-\bar{w}z} = \int_D \frac{\bar{w} \, d\mu(w)}{(1-\bar{w}z)^2} = P_{L^2_a}(\bar{w} \mu),$ which is (formally) the $L^2_a(D)$ Bergman space projection operator applied to the measure $\bar{w} \mu$; but I don't know if this function of $z$ necessarily does lie in $L^2_a$.</p> <p>So it's closely connected to Bergman space Toeplitz operators, which are known to be very tricky (much worse than on the Hardy space). Even worse, there is usually no nice way to characterise various function spaces which arise in terms of Taylor coefficients.</p> <p>So your question really combines <em>two</em> interesting and highly non-trivial problems! I doubt there is any really simple answer.</p> http://mathoverflow.net/questions/35449/why-isnt-the-theorem-of-approximation-applicable-in-banach-spaces/35454#35454 Answer by Zen Harper for Why isn't the theorem of approximation applicable in Banach spaces? Zen Harper 2010-08-13T09:11:39Z 2010-08-16T02:30:56Z <p>EDIT: since I'm not an expert on Banach spaces, I feel I shouldn't say anything more, but anyway; an essential ingredient is an <strong>exact</strong> formula in Hilbert spaces for $\|x + y\|^2$. <strong><em>Just an idea</em></strong> (<em>perhaps I am being stupid</em>): maybe if you have a Banach space where $\| x + y \| = F(\|x\|, \|y \|, g(x,y))$ for some reasonably simple functions $F$, $g$ then something can be said.</p> <p>If you examine the proof for Hilbert spaces, it makes essential use of the scalar product; so it's not really surprising that it doesn't work for general Banach spaces. The norm is a nice enough structure to do a lot of things, but not <em>that</em> nice.</p> <p>It also demonstrates that Banach spaces have far more detailed structure than just ordinary vector spaces, but that Hilbert spaces have even more structure again. In fact, there are many properties Hilbert spaces have which general Banach spaces don't (and many which even characterise Hilbert space uniquely).</p> http://mathoverflow.net/questions/34763/continuous-holomorphic-on-a-dense-open-holomorphic/34922#34922 Answer by Zen Harper for Continuous + holomorphic on a dense open => holomorphic? Zen Harper 2010-08-08T13:31:38Z 2010-08-08T13:31:38Z <p>Not an answer, but too big for a comment and useful for many similar problems.</p> <p>From Chapter VI of <strong><em>Theory of the Integral</em></strong> by Stanislaw Saks, p.197, available online at <a href="http://banach.univ.gda.pl/pdf/saks/" rel="nofollow">http://banach.univ.gda.pl/pdf/saks/</a></p> <p>For a domain $G$:</p> <p><strong>Theorem</strong> <em><strong>(Attributed to Besicovitch)</em></strong> Let $f$ be continuous on $G$, satisfying:</p> <p>(i): $f'(z)$ exists for each $z \in G \setminus E_1$,</p> <p>(ii): $$\limsup_{h \to 0} \left| \frac{f(z+h) - f(z)}{h} \right| &lt; \infty$$<br> for all $z \in G \setminus E_2$.</p> <p>(iii): $E_1$ has Lebesgue measure zero. </p> <p>(iv): $E_2$ is a countable union of finite-length curves. </p> <p><em><strong>Then</em></strong>: $f$ is holomorphic on $G$.</p> http://mathoverflow.net/questions/26086/analytic-continuations/34552#34552 Answer by Zen Harper for analytic continuations Zen Harper 2010-08-04T19:50:13Z 2010-08-04T19:57:39Z <p>Hi,</p> <p>I would say analytic continuation is, in many cases, almost like black magic!</p> <p>These problems are very, very hard. There is probably no sensible general answer. Not surprisingly, since solving the Riemann Hypothesis is trivially equivalent to the question of whether $1/\zeta$ has a continuation to $\{ \mathrm{Re}(z) > 1/2 \}$ or not!</p> <p>Analytic continuation is very ill-posed: approximating a function $F$ by a sequence $F_n$ usually is of no use at all in determining the domain you can extend $F$ to. So you usually need <strong><em>exact formulae</em></strong> for your functions, expressed as infinite series, products, integrals or something else; <strong><em>numerical computations are almost certainly useless</em></strong> for these problems. </p> <p>Often, existence of a continuation depends on <em>cancellation</em> in very complicated oscillating series or integrals, so you rarely have nice things like <em>absolute</em> convergence (or even conditional convergence!) on the true domain of the function. It can be very difficult to analyse the formulae you get.</p> <p>Unfortunately you probably will have to do lots of detailed algebraic calculations with the specific functions you are considering. Cauchy's theorem for functions expressed using contour integration is about the only general method I can think of, although it's not always possible.</p> http://mathoverflow.net/questions/33911/why-linear-algebra-is-funor/34068#34068 Answer by Zen Harper for Why linear algebra is fun!(or ?) Zen Harper 2010-07-31T23:59:35Z 2010-07-31T23:59:35Z <p>A bit rubbish and easy, but amusing if you haven't seen it before.</p> <p>Let $G$ be a finite group such that $g^2=e$ for all $g \in G$, i.e. every element (<em>except the identity $e$</em>) has order 2. Then $G$ has size $2^n$ for some $n$.</p> <p>This is not too hard to prove directly; but it becomes totally obvious (<em>once you've proved that $G$ is abelian</em>) when you realise that $G$ is a finite-dimensional vector space over $F_2$.</p> http://mathoverflow.net/questions/31458/on-linear-independence-of-exponentials/33023#33023 Answer by Zen Harper for On linear independence of exponentials Zen Harper 2010-07-22T23:41:51Z 2010-07-22T23:46:56Z <p>Still thinking about the interesting question!</p> <p><em><strong>Not an answer</em></strong>, but too big for a comment.</p> <p>To show what I meant in my comment to Daniel Litt's answer about the difference between uniform <em>absolute</em> convergence and ordinary uniform convergence:</p> <p>I <strong>think</strong> (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly on the set $\{ z \in \mathbb{C} : |z| \leq 1 \}$, but not uniformly absolutely. </p> <p>Thus $\sum_{n=0}^\infty |u_n| = +\infty$, but also $$\forall \, \epsilon>0, \quad \exists \, N(\epsilon)>0 \quad \text{such that}: \qquad \forall \, |z| \leq 1, \quad \forall \, m \geq n > N(\epsilon), \qquad \left| \sum_{k=n}^{m} u_k z^k \right| &lt; \epsilon.$$ This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies $$f \in A(D) \setminus W_+(D),$$ which shows in particular that the <em>disc algebra</em> $A(D)$, consisting of functions continuous on the closed unit disc and analytic on the open unit disc, is <strong>strictly larger</strong> than the <em>Wiener algebra</em> $W_+(D)$ of power series absolutely convergent on the closed unit disc.</p> <p>So the problem is going to be pretty hard because we can't use <strong>absolute</strong> convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).</p> <p>The easiest "solution" is just to ignore it (i.e. <strong><em>assume</em></strong> absolute convergence!) This gives a slightly different, but still interesting, question.</p> http://mathoverflow.net/questions/29499/yet-another-roadmap-style-request-a-second-bite-of-the-cherry/32575#32575 Answer by Zen Harper for Yet another 'roadmap' style request- a second bite of the cherry Zen Harper 2010-07-20T01:41:55Z 2010-07-20T01:41:55Z <p>???!!!</p> <p>My answer is similar to Justin Curry's answer, except that I am specifically talking about the UK, and advocating a more formal approach.</p> <p>Speaking as someone who <em>does</em> have a Ph.D. from England, and knows people who were given Ph.D. funding from EPSRC despite relatively poor first degrees (and also knowing someone who was given 3 years' funding to start a Ph.D. a <em>second</em> time, after dropping out of his first Ph.D. in the first year - EPSRC may be more generous than you think, although it's probably getting worse):</p> <p>I believe your plan is totally wrong and worse than useless. The best that can happen is you spend one year learning a bit of stuff, and rediscover a few known results (which won't count for much; rediscovery is useless for funding purposes, unless your method is totally different and better than known ones). Far more likely, you will waste another year.</p> <p>The first thing is, no matter how much you might think you know now, you must <strong><em>know that you know NOTHING!</em></strong> Without proper, regular guidance from a research supervisor, your plan is <strong><em>GUARANTEED</em></strong> to fail, unless you're a genius or very lucky (<em>preferably both</em>). You simply <strong><em>CANNOT</em></strong> do good, worthwhile, <strong><em>original</em></strong> research yourself without a supervisor. You should not even be talking about research at all now, that's what Ph.D.s are for!</p> <p>Assuming you want to stay in the UK, as far as I can see the answer is simple. </p> <p>A 2:1 from Warwick is not that bad.</p> <p>Your best chance is to do a one-year M.Sc. (or equivalent) at the best university you can (try, e.g., the Cambridge Part III, which has different funding rules from most others). If it goes well, funding for a Ph.D. will be no problem. If it goes badly, well, ask the question again next year...!</p> <p>A good M.Sc. will more than make up for your 2:1 (<em>which, as I said, is not so bad; a 2:2, on the other hand, would leave you in a very tricky situation!</em>)</p> <p>You should do this <strong><em>IMMEDIATELY!!!</em></strong> You still should have enough time to enter in September 2010 if you hurry. </p> <p>If you can't get funding to do the M.Sc./Part III or whatever, you'll just have to borrow money and pay for it yourself with student loans. If you're not willing to do this, as a <strong><em>LAST RESORT</em></strong> you should live near a good university and sneak into the lectures without registering with the university. This is (<em>I believe</em>) perfectly legal, since UK university lectures are still open to the public, as long as you don't disturb anyone. You will almost certainly be unnoticed in Oxford or Cambridge (since everyone will just assume you're from a different College), or large universities where many students don't know each other. </p> <p>If you ask the Maths Department very nicely, they'll probably let you use their library (and if you pay, they definitely will). But they almost certainly won't let you sit any exams or have any help/tuition unless you pay. You could make private tuition arrangements with Ph.D. students or even university staff; this would probably be much less expensive than registering properly with the university, if you only have a small number of specific courses to look at.</p> <p>Like I said, doing some M.Sc. lectures without the actual final qualification is a very poor solution, but it's still better than your proposed plan.</p> http://mathoverflow.net/questions/28158/a-learning-roadmap-request-from-high-school-to-mid-undergraduate-studies/32251#32251 Answer by Zen Harper for A Learning Roadmap request: From high-school to mid-undergraduate studies Zen Harper 2010-07-17T02:45:08Z 2010-07-19T17:00:58Z <p><em><strong>EDIT:</em></strong> <em>some people didn't like what I've written below. Rather than change it, I've added comments below instead. Why should I change it anyway? If you don't agree, that's your opinion; but these are mine, so why shouldn't Max see a fair range of differing views? One of the things I was trying to say is that if you want to have a successful career ACTUALLY USING PROPER MATHEMATICS beyond just, say, calculating percentages or putting numbers into a formula, then your chances really are very slim unless you're EXCEPTIONALLY good. The number of good jobs with genuine mathematical content is far, far smaller than the number of mathematicians with reasonable ability (at least in England). The sooner Max knows this, the better he can prepare properly and take precautions.</em></p> <p>Hi Max, it's nice to see some youngsters with interest!</p> <p>Firstly, Hardy's <em>Divergent Series</em> is a great, fun book, but probably too hard and of little use for your future career (<em>it's very old-fashioned and not likely ever to come back into fashion - but I could be wrong</em>).</p> <p>My personal opinions for what they're worth (i.e., not very much); some (<em>many?</em>) people would disagree!</p> <p><em><strong>The most important thing</em></strong> by far is to go to the very best university you can. I assume that you have genuine mathematical talent (<em>you must be easily the best at mathematics in your school</em>) and that you want to go to the top university in Holland; don't waste your time at second best places. (<em>If not, then improving your school grades or whatever you need is the top priority!</em>) It's far better to be in the middle at the best university than at the top in a poor university. </p> <p>If you can't decide which is the "best", ask your teachers, see which ones have Fields Medallists/Nobel Prize winners, look at league tables, etc. (<em>I'm English and don't know how the Dutch system works, so apologies if this sounds like nonsense</em>). As an absolute minimum, it MUST be a university which offers Ph.D. degrees in all areas of mathematics.</p> <p>At good universities, you have nothing to worry about; your lectures/university library will provide everything you need.</p> <p>Almost certainly, you will buy the wrong books (<em>either far too easy or far too difficult</em>) that you'll never read (<em>or will quickly finish and want to throw away</em>). Almost certainly, your mathematical tastes will change significantly over the next few years.</p> <p>Sad, but true: many (<em>perhaps even "most" or "almost all"</em>) maths books are boring, tedious, unsuitable and impossible to read (<em>and different people will disagree about which books are which, so advice is not much use!</em>)</p> <p>Personally, almost every mathematical book I bought as an undergraduate or before was a total waste of money; with more experience you'll know better in future what to look for; but <strong><em>now is the worst possible time for you to buy books.</em></strong></p> <p><em><strong>Until you have spent at least one or two years at university, SAVE YOUR MONEY!!! GO TO LIBRARIES!! DO NOT BUY MATHEMATICS BOOKS!!!!!</em></strong> (<em>Unless you are rich or someone else is paying...</em>)</p> <p><em><strong>If you do buy books, only do so after you have looked at them for several weeks or months from a library!</em></strong></p> <p><em><strong>Only buy books with LOTS of exercises!</em></strong>(Apart from rare exceptions, e.g. Hardy and Wright: An Introduction to the Theory of Numbers.)</p> <p><em><strong>Doing fun research problems of your own, either directly or indirectly inspired by books,</em></strong> is just as important as pure reading; and try not to read too much right now. You won't discover anything new, but that's not the purpose (and the internet is <em>too</em> big - don't look up the answers too quickly, or you'll never develop).</p> <p><strong>Don't listen to everything physicists say when they're talking about mathematics...!</strong> Physicists and mathematicians are very different.</p> <p><strong>Practice basic calculations: calculus, infinite series, Fourier series, complex numbers, differential equations, basic number theory,... and whatever else interests you.</strong> It will be very useful in future.</p> <p><strong>Try looking through things on Wikipedia and the links, and just following wherever your interests take you.</strong> You'll learn a surprising amount this way, and enjoy it more too. (<em>But be very careful of the Internet; almost everything online, if not written by proper university mathematicians, is rubbish! MathOverflow and Wikipedia are rare exceptions, but even these have a small amount of rubbish on them</em>).</p> <p>And finally...</p> <p><em><strong>BE FLEXIBLE!</em></strong> Many things you are interested in now might not be to your taste in the future (<em>mathematics is very, very large and no-one can do all parts of it</em>); it is even possible (<em>what a hideous thought!</em>) that you might lose your interest in Mathematics and prefer Physics, Computer Science, Engineering or something else instead; be prepared to change! (<em>Very few people survive up to Ph.D. level and beyond; if you don't become a mathematician, you will get much better jobs after graduation if you do these subjects at university instead...!</em>)</p> <p>If you still want to study mathematics at university: <strong>be prepared for lifetime social exclusion, poverty and unemployment!</strong> (This is more for Ph.D. than B.Sc.; perhaps I exaggerate slightly, but you must prepare for the dangers!)</p> http://mathoverflow.net/questions/32225/slap-your-forehead-moments-greatest-hits/32254#32254 Answer by Zen Harper for "Slap your forehead" moments- Greatest Hits Zen Harper 2010-07-17T03:07:23Z 2010-07-17T03:07:23Z <p>For some problem in Algebraic Topology (presumably related to homotopy groups or similar, with free groups): I thought that two groups $G$, $H$ were isomorphic, because $G \approx A \subseteq H \approx B \subseteq G$, where $\approx$ means "<em>is isomorphic to</em>" and $\subseteq$ means "<em>is a subgroup of</em>".</p> <p>However, I was very shocked when informed that this does NOT imply that $G \approx H$ in general! (<em>I thought this was true, and spent a long time trying to prove it; but I knew that I hadn't succeeded. So I suppose this doesn't qualify, but anyway</em>).</p> http://mathoverflow.net/questions/31308/splines-harmonic-analysis-singular-integrals/31318#31318 Answer by Zen Harper for Splines, harmonic analysis, singular integrals. Zen Harper 2010-07-10T18:41:49Z 2010-07-10T18:41:49Z <p>If you want to extend <strong>differentiation</strong> to all continuous functions, then (provided you have some convenient mathematical properties of the extension) you are FORCED to use distributions or roughly equivalent things; you have no choice! Similarly, to extend <strong>the Fourier transform</strong> you are forced to consider <em>tempered distributions</em>.</p> <p>Speaking as a pure mathematician: the main purpose of general distributions is to extend <em>differentiation</em>, not integration (since integration makes things nicer; it is differentiation which is the nastier operation). They are fine as long as you aren't using the Fourier transform.</p> <p>Thus, every locally integrable function can be regarded as a distribution, and therefore differentiated; so, when you're considering differential equations, this might be all you need (you don't have to worry whether the functions are differentiable or not, because distributions always are). You find distribution solutions, then try to prove that they're actually functions.</p> <p>It's similar to solving polynomial equations by using complex numbers; even if all the roots are real, it's still sometimes easier to solve them with complex numbers, then try to prove they're real (e.g. by showing they're self-conjugate).</p> <p>However, if you want to do <em>Fourier Transforms</em> then you have to consider <em>tempered distributions</em> (or Schwartz distributions), since general distributions are sometimes too nasty to have Fourier transforms.</p> <p>Note that even genuine locally integrable functions need not represent tempered distributions, so general distributions are not appropriate for Fourier transforms even when you only want to consider functions.</p> <p>But <em>Fourier inversion</em> works perfectly for tempered distributions, no further restrictions are needed, unlike, say, $L^1$. If $f \in L^1$ then $\widehat{f}$ is usually not in $L^1$, so you can't do Fourier inversion theory nicely on $L^1$ (you would have to <strong>assume</strong> that also $\widehat{f} \in L^1$, which is often not true!)</p> <p>Extension in mathematics is very powerful; when you don't have to worry about restrictions and annoying details, it is easier! For example, <strong>complex numbers are easier than real numbers</strong>, <strong>complex analysis is easier than real analysis</strong>, and <strong>Lebesgue integration is easier than Riemann integration</strong>!! Students never believe this, but it's true if you actually want to <strong>use</strong> it (rather than do toy problems in books)...</p> http://mathoverflow.net/questions/31295/let-a-function-f-have-all-moments-zero-what-conditions-force-f-to-be-identically Let a function f have all moments zero. What conditions force f to be identically zero? Zen Harper 2010-07-10T13:34:50Z 2010-07-10T18:04:33Z <p>Throughout, let $f$ be a Lebesgue measurable function (or continuous if you wish, but this is probably no easier). (<em>Questions with distributions etc. are possible also but I want to keep things simple here</em>).</p> <h3>FINAL CLARIFICATION/REWRITE!!</h3> <p>Thanks to all who have commented so far. I will need some more time to digest it properly. The original forms of the questions are at the end; here I have rewritten the questions, hopefully more clearly; sorry for my poor explanation before!</p> <p><strong>Definition</strong> $\mathcal{M}_a$, the space of <em>absolutely null moment functions</em>, is the set of all measurable $f$ satisfying $$\int_0^\infty t^n |f(t)| dt &lt; \infty, \quad \int_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots.$$ $\mathcal{M}$, the space of <em>null moment functions</em>, is the set of all measurable $f$ satisfying $$\int_0^T |f(t)| dt &lt; \infty \quad \forall \, T>0, \qquad \lim_{R \to \infty} \int_0^R t^n f(t) dt = 0 \quad \forall \, n=0,1,2,\ldots$$</p> <p>Thus, trivially $0 \in \mathcal{M}_a \subseteq \mathcal{M}$, but $\mathcal{M}_a$ contains many other non-trivial functions. It seems certain that $\mathcal{M}_a \ne \mathcal{M}$ (I would be amazed if the spaces were equal), although constructing an explicit example seems tricky.</p> <p><strong>Definition</strong> Given a function $\psi \geq 0$, let $G(\psi)$ be the set of all $f$ such that $|f| \leq \psi$.</p> <p>(<em>Of course we're identifying functions equal a.e., so really we should consider equivalence classes etc. just as for</em> $L^p$ <em>spaces</em>).</p> <p><em><strong>Rephrased Question 1</em></strong> Find general simple necessary and/or sufficient conditions on $\psi$ with the property $$G(\psi) \cap \mathcal{M}_a = \{ 0 \}.$$</p> <p><em><strong>Rephrased Question 2</em></strong> The same as Question 1, but with $G(\psi) \cap \mathcal{M}$ instead.</p> <p>Or, if $G(\psi)$ is not the appropriate space for these problems, consider $\int_T^{2T} |f| dt \leq g(T)$ or $\int_n^{n+1} |f| dt \leq A_n$ or something similar instead. Finding the correct kind of restrictions on $f$ is part of the problem.</p> <p>Thus, $G(\psi) \cap \mathcal{M}_a = \{ 0 \}$ for $\psi(t) = \exp(-\delta t)$, by the discussion below; and also for any compactly supported $\psi \in L^1$. </p> <p>But $G(\psi) \cap \mathcal{M}_a \ne \{ 0 \}$ for $\psi(t) = \exp(-t^{1/4}) |\sin(t^{1/4})|$.</p> <hr> <h3>Original QUESTION 1</h3> <p>If $$\int_0^\infty t^n |f(t)| dt &lt; \infty, \quad \int_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots,$$ when must $f \equiv 0$ almost everywhere? </p> <hr> <p><em><strong>EDIT: CLARIFICATION</em></strong>: this is really about <em>classes of functions, expressed in terms of a growth/decay rate function</em> $\phi$, <em>which give</em> <strong>unique</strong> <em>solutions to the moment problem</em>. <strong><em>I am NOT asking</em></strong> how to solve the moment problem itself! </p> <p>If possible, find <em>necessary and sufficient conditions</em> on functions $\phi \searrow 0$ with the property that $$\int_R^\infty |f(t)| dt \leq \phi(R) \qquad \Rightarrow \qquad f \equiv 0.$$ So, $\phi(R) = \int_R^\infty \exp(-t^{1/4}) |\sin(t^{1/4})| dt$ is <em>not enough</em> (by Example 2 below); nor is $\phi(R) = \int_R^\infty |f(t)| dt$ with f given by "coudy" in his/her answer below.</p> <p>But $\phi = \chi_{[0,b]}$ is enough (by Example 1 below); moreover $\phi(R) = \exp(-\delta R)$ would be enough for any fixed $\delta > 0$, by my discussion of Example 1 below, because the relevant Laplace transform $F = \mathcal{L}f$ is analytic on the half-plane $\{ \mathrm{Re}(z) > -\delta \}$. So we want to know about the gap <em>between</em> $\exp(-\delta R)$ and functions like that given by "coudy" below.</p> <p><em><strong>FURTHER CLARIFICATION:</em></strong> by <strong>analogy</strong>, maybe an example from PDEs will explain better (I'm not saying this is related to my problem; I'm saying that this is <em>the same kind of result</em> as what I want):</p> <p><strong>Definition</strong> A <em>null temperature function</em> is a continuous function $u = u(x,t) : \mathbb{R} \times [0, \infty) \to \mathbb{R}$ such that the <em>heat equation</em> is satisfied, i.e. $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ in $\{ t>0 \}$, and $u(x, 0) = 0$ for all $x$.</p> <p><strong>Theorem</strong> There exists a null temperature function satisfying $|u(x,t)| \leq \exp(A/t)$ with $A>0$, such that $u(x,t) \not\equiv 0$ for all $t>0$.</p> <p><strong>Theorem</strong> Let $u$ be a null temperature function satisfying $|u(x,t)| \leq A \exp(B t^{-\delta})$, for some $A,B>0$ and $\delta&lt;1$. Then $u \equiv 0$.</p> <p>So here the "critical" growth rate for null temperature functions is roughly $\exp(A/t)$. I am looking for a similar thing with "null moment functions".</p> <p>Note that this is a <strong><em>totally different problem</em></strong> to: <em>given</em> $v$, <em>find some</em> $u$ satisfying the heat equation such that $u(x,0)=v(x)$.</p> <hr> <h3>Original QUESTION 2</h3> <p>If instead we have only $\int_0^R |f(t)| dt &lt; \infty$ for each $R>0$, and $$\lim_{R \to \infty} \int_0^R t^n f(t) dt = 0, \qquad n=0,1,2,\ldots$$ when must $f \equiv 0$ almost everywhere? <em>(I have very little idea about this)</em>.</p> <hr> <p>I think these questions are clearly very natural, interesting, and important, but Googling etc. didn't work well (I tried "vanishing moments" and other phrases, but there's just too much stuff out there). Standard known examples/methods follow.</p> <p><em><strong>Example 1:</em></strong> if $f$ is compactly supported on $[a,b]$, say, then $f \equiv 0$ a.e. because polynomials are dense in $C[a,b]$.</p> <p><em><strong>Example 2:</em></strong> by taking imaginary parts of $\int_0^\infty t^{4n+3} \exp(-(1+i)t) dt \in \mathbb{R}$, we get $$\int_0^\infty t^{4n+3} e^{-t} \sin t \, dt = 0,$$ and so by substituting $t = x^{1/4}$, $$\int_0^\infty x^n \exp(-x^{1/4}) \sin(x^{1/4}) dx = 0, \qquad n=0,1,2,\ldots$$</p> <p><em>Alternative method for Example 1:</em> consider the Laplace transform $F(z) = \int_0^\infty e^{-zt} f(t) dt$. In Example 1, $F$ is an entire function such that $F^{(n)}(0) = 0$ for all $n$, so $F \equiv 0$ and thus $f \equiv 0$ a.e. as required.</p> <p>So, any condition on $f$ forcing $F$ to be analytic on some disc with centre $0$ is enough; but can we do better? </p> <p>In Example 2, $f \in L^1(0,\infty)$ and so $F$ is bounded and analytic on $\{ \mathrm{Re}(z) > 0 \}$, and continuous on the boundary, with $\lim_{z \to 0} F^{(n)}(z) = 0$ for all $n$. But this is still not enough to force $F \equiv 0$.</p> http://mathoverflow.net/questions/30156/demystifying-complex-numbers/30980#30980 Answer by Zen Harper for Demystifying complex numbers Zen Harper 2010-07-08T00:52:14Z 2010-07-08T00:52:14Z <p>Maybe artificial, but a nice example (<em>I think</em>) demonstrating <em>analytic continuation</em> (<strong><em>NOT</em></strong> just the usual $\mathrm{Re}(e^{i \theta})$ method!) I don't know any reasonable way of doing this by real methods.</p> <p>As a fun exercise, calculate $$I(\omega) = \int_0^\infty e^{-x} \cos (\omega x) \frac{dx}{\sqrt{x}}, \qquad \omega \in \mathbb{R}$$ from the real part of $F(1+i \omega)$, where $$F(k) = \int_0^\infty e^{-kx} \frac{dx}{\sqrt{x}}, \qquad \mathrm{Re}(k)>0$$ (<em>which is easily obtained for $k>0$ by a real substitution</em>) and using analytic continuation to justify the same formula with $k=1+i \omega$.</p> <p>You need care with <em>square roots</em>, <em>branch cuts</em>, etc.; but this can be avoided by considering $F(k)^2$, $I(\omega)^2$.</p> <p>Of course all the standard integrals provide endless fun examples! (But the books don't have many requiring <em>genuine</em> analytic continuation like this!)</p> http://mathoverflow.net/questions/19997/if-the-fourier-transformed-of-f-is-odd-is-f-odd/29077#29077 Answer by Zen Harper for If the fourier transformed of f is odd, is f odd? Zen Harper 2010-06-22T11:25:10Z 2010-06-22T11:25:10Z <p>Yes. Using <em>tempered distributions</em> it's immediately obvious, since $f = C \mathcal{F}^3 (\mathcal{F} f)$ for a constant $C$, and $\mathcal{F}$ maps odd <em>distributions</em> into odd <em>distributions</em>.</p> <p>The missing details of the proof are just simple exercises in distributions.</p> <p>Distribution theory is very useful for Fourier transform questions like this, since Fourier inversion works perfectly.</p> http://mathoverflow.net/questions/107945/nonstandard-analysis-in-probability-theory/107959#107959 Comment by Zen Harper Zen Harper 2012-09-27T01:35:37Z 2012-09-27T01:35:37Z But doesn't nonstandard analysis <i>need</i> the Axiom of Choice, or at least need more than just ZF? Of course, some (many?) people include AC within &quot;standard analysis&quot;, especially if they include functional analysis, but all the basic &quot;standard analysis&quot; stuff used to do real variable calculus doesn't use full AC, only a weak axiom of countable dependent choice or similar. http://mathoverflow.net/questions/108046/is-it-common-practice-to-publish-parts-of-a-phd-thesis-in-advance Comment by Zen Harper Zen Harper 2012-09-25T13:47:54Z 2012-09-25T13:47:54Z But, being cynical, I believe that, with rare exceptions, almost no-one reads Ph.D. theses apart from the examiners <i>if you're lucky</i> ; published papers are much more important because that's what far more people will actually read and cite <i>if you're lucky</i> . But choose your journals with care! http://mathoverflow.net/questions/108046/is-it-common-practice-to-publish-parts-of-a-phd-thesis-in-advance Comment by Zen Harper Zen Harper 2012-09-25T13:44:05Z 2012-09-25T13:44:05Z The style of Ph.D. theses and research papers is different, so I think it's perfectly normal (and even encouraged) to submit papers first, and rewrite/expand them for the thesis. A thesis should contain lots of extra background and trivial details which you don't put in a paper, and it's far easier and more interesting to write papers first and then convert/expand them into thesis chapters than the other way round; and, once converted into thesis format, they will be sufficiently different that copyright could not be an issue. But be sure to acknowledge all this in the papers and the thesis. http://mathoverflow.net/questions/106026/non-analyticity-of-convolution Comment by Zen Harper Zen Harper 2012-09-02T22:42:21Z 2012-09-02T22:42:21Z Have you tried using a Taylor series expansion in $(z^2-1)^{1/2} = z(1-1/z^2)^{1/2}$, for $z = |x-y|&gt;1$? Then, at least formally, you can get a decomposition into simpler operators, using a formal infinite sum. http://mathoverflow.net/questions/106026/non-analyticity-of-convolution Comment by Zen Harper Zen Harper 2012-09-02T22:38:52Z 2012-09-02T22:38:52Z I think it seems unlikely, without a lot of extra assumptions, that there will be any useful general statement. Knowing that g is not analytic at some point doesn't really give you much information, since there are many possible ways for analyticity to fail; and furthermore, one single value of g (maybe even all its derivatives) doesn't change f in any way (whereas by contrast, having g analytic at a point tells you something about g on a whole interval). Of course if you could invert the integral operator and express g directly in terms of f, you would get a lot of information. http://mathoverflow.net/questions/103453/does-anybody-know-an-estimation-of-l4-norm-of-fejer-kernel Comment by Zen Harper Zen Harper 2012-07-29T15:05:06Z 2012-07-29T15:05:06Z I agree with Davide Giraudo, I think the Fejer kernel has a square, so you are either missing a square, or using the wrong name. Have you tried just expressing with complex exponentials and evaluating directly? Since you're raising a trig. polynomial to a positive integer power, it has an exact closed form expression, which may be easier to estimate. http://mathoverflow.net/questions/103014/lacunary-sequence Comment by Zen Harper Zen Harper 2012-07-24T16:12:11Z 2012-07-24T16:12:11Z If there is a standard definition, it is clearly not well-known enough to be used without stating it; I think I've seen similar things with lim inf instead of lim. For a paper, it's safest to define exactly what you mean. http://mathoverflow.net/questions/98671/an-inverse-laplace-transform-involving-error-function Comment by Zen Harper Zen Harper 2012-06-04T09:14:59Z 2012-06-04T09:14:59Z Sorry, I meant &quot;convolve&quot; rather than &quot;multiply&quot;, of course (Laplace transforms change convolution into multiplication). http://mathoverflow.net/questions/86760/on-the-generalisation-of-the-laplace-transform Comment by Zen Harper Zen Harper 2012-06-04T09:12:26Z 2012-06-04T09:12:26Z I think this is too general to have any useful answer, without extra conditions on $g$; although I'd be glad to be wrong, since any nontrivial general results on this would be very interesting. If you know $g$ satisfies a differential equation, for example, you could start to do stuff. http://mathoverflow.net/questions/98671/an-inverse-laplace-transform-involving-error-function Comment by Zen Harper Zen Harper 2012-06-04T09:05:58Z 2012-06-04T09:05:58Z If you know the inverse transform of $g$, then you can get $g_a$ by a convolution, since $g_a(z)= (1-2 / \sqrt{z})^{-1} g(z)$, so you only need to find the inverse transform of $(1-2 / \sqrt{z})^{-1}$, which is an easy power series in $1/ \sqrt{z}$, then multiply. There's no guarantee that the resulting expression is particularly nice or useful, though. http://mathoverflow.net/questions/98451/understanding-the-inverse-laplace-transform-of-a-function-with-essential-singular Comment by Zen Harper Zen Harper 2012-06-01T10:41:03Z 2012-06-01T10:41:03Z I've added the Fourier Analysis tag, since I think it is appropriate [although that usually goes without saying, when Laplace transforms are used!] http://mathoverflow.net/questions/98451/understanding-the-inverse-laplace-transform-of-a-function-with-essential-singular Comment by Zen Harper Zen Harper 2012-05-31T11:20:58Z 2012-05-31T11:20:58Z Hi, this question will probably be closed as not being &quot;research level&quot;. But I would say: (1) what is the exact function you want to invert? You haven't said in your question; (2) Inverse Laplace transforms don't always exist, you need conditions on the analytic function; (3) The Residue Theorem works for essential singularities also, only the coefficient of 1/z is important. Whether the negative powers of z are finite or infinite in number is irrelevant, since the Laurent series always converges nicely enough to interchange contour integration with the infinite sum. http://mathoverflow.net/questions/92969/harmonic-function-and-mvp Comment by Zen Harper Zen Harper 2012-04-03T16:57:19Z 2012-04-03T16:57:19Z Also, assuming $f$ is continuous, then if we fix the centre $x$ and let $MV(\delta)$ be the Mean Value of $f$ over the ball of radius $\delta &gt; 0$ centred at $x$, then $MV$ is a continuous function of $\delta$ by Dominated Convergence (or just uniform continuity of $f$), so you can assume the set of radii is <i>closed</i>. But in fact, the case $f$ continuous is apparently already partially solved, according to the MO question and answers referred to in Andrey Rekalo's above comment. http://mathoverflow.net/questions/92969/harmonic-function-and-mvp Comment by Zen Harper Zen Harper 2012-04-03T16:46:53Z 2012-04-03T16:46:53Z You probably need some extra condition on $f$ to say that MVP implies Harmonic; say, $f$ continuous. (<i>At the very least, you'd need Lebesgue measurability etc. even to define the Mean Value as an integral!</i>) Assuming $f$ continuous, <i>local</i> MVP is enough: for each point $x$ there exists $\delta(x) &gt; 0$ such that MVP holds on all spheres centred at $x$ of radius $&lt; \delta(x)$. But for example, $f(x,y) = \sgn(x)$ has the local MVP on $R^2$, but is not continuous. http://mathoverflow.net/questions/85065/unexpected-applications-of-the-fact-that-nth-degree-polynomimals-are-determined-b/85102#85102 Comment by Zen Harper Zen Harper 2012-01-07T03:44:24Z 2012-01-07T03:44:24Z Also really nice is that you can prove $s(n)$ is always divisible by $n(n+1)$, and by $2n+1$ for $k$ even, and by $n^2(n+1)^2$ for $n&gt;1$ odd, by considering $s(-n)$ and using induction, thus getting formulae for $s(-x)$ in terms of $s(x)$, and then considering zeros of $s$ and its <i>derivative</i> $s'$. I still love the fact that values at non-integers, and derivatives, can be relevant to a question originally only about <i>positive</i> integers. This is also motivation for how the Riemann Zeta Function and analytic continuation can tell us about prime numbers.