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Nov
20 |
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What is the difference between hard and soft analysis?
I'm not exactly sure what you mean that this is a probabilistic argument. One can phrase it in a probabilistic way, but really it boils down to a hard linear algebra problem since the Markov process has finitely many states (the measures here are just vectors/functions). The statement that convergence happens is just the Perron-Frobenius Theorem. It is rather convenient to use probabilistic methods in the hard analysis of shuffling, though. I think this example rather clearly demonstrates the struggle between generality and quantitativeness. By the way, I myself happen to be an analyst... |
Nov
20 |
answered | What is the difference between hard and soft analysis? |
Nov
3 |
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What's your favorite equation, formula, identity or inequality?
oops, here is the link: cornellmath.wordpress.com/2008/02/15/… |
Nov
3 |
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What's your favorite equation, formula, identity or inequality?
I am a bit late in seeing this post, but I completely agree with you, Yaakov. The equality is a bit related to the following pattern which I discovered as a child but continues to amaze me to this day: $1=1^3$; $3+5=2^3$; $7+9+11=3^3$; $13+15+17+19=4^3$;... Many mathematicians know that the sum of the first n odd numbers is n2, but I think very few are aware of this trivial yet incredible pattern. I actually wrote a little post about this on a math blog (the Everything Seminar): |
Oct
6 |
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The Cauchyâ€“Riemann equations and analyticity
Your second comment is absolutely spot on, Nate. |
Sep
29 |
awarded | Nice Answer |
Sep
29 |
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Ingenuity in mathematics
I completely agree that it is not hard to convince people of the answer (especially when you explain visually what's happening with some props), but I have never met anyone who instinctively would answer that it's better to switch. The result is extremely counterintuitive to essentially everyone, well-educated or not. It's the kind of problem where most people will struggle and disagree with you at first until they have a eureka moment and it suddenly seems so obvious. |
Sep
29 |
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Ingenuity in mathematics
This is indeed a cool fact. However, I am certain any non-math person would be lost at the term matrix, let alone diagonalize, especially when said non-math people merely suppose some mathematical proofs exhibit ingenuity :) |
Sep
28 |
answered | Ingenuity in mathematics |
Sep
22 |
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What is the naming reason of poles in complex analysis?
According to the online etymological dictionary, the English word pole derives from the Latin word palus meaning stake. So the term could be handed down to us from when mathematicians still wrote in Latin. The term could also derive from Greek, polos (axis of a sphere), or it could be a coincidence. Simply because the direct cognate of a word in German doesn't correspond to the current meaning of an English word doesn't mean the words didn't have similar meanings long ago. The English word table is related to the German word Tafel even though Tafel only infrequently means table. |
Sep
22 |
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What is the naming reason of poles in complex analysis?
@Martin: According to the online etymological dictionary, the English word pole derives from the Latin word palus meaning stake. So the term could be handed down to us from when mathematicians still wrote in Latin. The term could also derive from Greek, polos (axis of a sphere), or it could be a coincidence. Simply because the direct cognate of a word in German doesn't correspond to the current meaning of an English word doesn't mean the words didn't have similar meanings long ago. The English word table is related to the German word Tafel even though Tafel only infrequently means table. |
Jul
22 |
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Spherical Harmonics - a bunch of questions about them
There are nice sections about spherical harmonics in two of Stein's books Singular Integrals and Differentiability Properties of Functions and Introduction to Fourier Analysis on Euclidean Spaces, but if you don't know anything about $L^2$, these treatments could very well be too difficult. |
Jul
18 |
answered | Why pi-systems and Dynkin/lambda systems? On the relative merits of approaches in measure theory. |
Jul
13 |
awarded | Enthusiast |
Jul
10 |
answered | Splines, harmonic analysis, singular integrals. |
Jul
10 |
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Splines, harmonic analysis, singular integrals.
Your statement about functions of compact support is the correct problem with defining the Fourier transform of distributions with smooth compactly supported functions as the test functions. |
Jul
10 |
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Splines, harmonic analysis, singular integrals.
These are good ideas, but I think your concluding sentence is a bit too strong. One could easily define the space of test functions to be the space of compactly supported smooth functions. The Fourier transform of such a function, being a very special kind of Schwarz function, is also smooth with extreme decay at infinity. The problem is that while any locally integrable function defines a linear functional over $C_c^{\infty}$, the formula $\hat{T}(\phi)=T(\hat{\phi})$ does not make sense since $\hat{\phi}$ will never be compactly supported when $\phi$ is compactly supported. |
Jul
2 |
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Demystifying complex numbers
I think your example with the Fourier expansion somewhat misses the fact that the Fourier transform is one of the most important mathematical tools in analysis and involves complex numbers. |
Jun
15 |
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Why do we care about $L^p$ spaces besides $p = 1$, $p = 2$, and $p = \infty$?
I think it's not that these spaces are the ones that matter but rather these spaces are the only ones where it is convenient to perform computations. Like I said in a comment to another answer, $L^2$ would be just as exotic if not for access to great Hilbert space tools. |
Jun
14 |
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Why do we care about $L^p$ spaces besides $p = 1$, $p = 2$, and $p = \infty$?
I tend to agree with you to some extent. With $L^\infty$ and $L^1$, things are much simpler computationally, and with $L^2$ you have the immense advantage of Hilbert space tools like Plancherel's identity. If not for Hilbert space techniques, one would likely find $L^2$ just as exotic as the others. The interpolation theorems basically say that understanding two of these spaces is enough to understand everything in between. So long as the adjoint of the operator is similarly behaved, just two is enough to understand all $p\ge 1$. This is the case with the Hilbert transform, for instance. |