most recent 30 from http://mathoverflow.net 2013-05-24T03:36:31Z http://mathoverflow.net/feeds/user/10843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51190/counterexample-markov-process/51192#51192 Michael Renardy 2011-01-05T10:07:18Z 2011-01-05T10:07:18Z

<p>The porous medium equation provides examples. See, for instance, M. Inoue, A Markov process associated with a porous medium equation, Proc. Japan Acad. 60 (1984), 157-160.</p>

http://mathoverflow.net/questions/44326/most-memorable-titles/50775#50775 Michael Renardy 2010-12-30T23:51:55Z 2010-12-30T23:51:55Z

<p>"A survey of finite differences of opinion on numerical muddling of the incomprehensible defective confusion equation" by B.P. Leonard</p>

http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product/50125#50125 Michael Renardy 2010-12-22T03:32:45Z 2010-12-22T03:32:45Z

<p>I think D was supposed to have positive entries. If B is positive definite (meaning that the associated quadratic form is positive definite), then so is $D^{1/2}BD^{1/2}$. This matrix is similar to $DB$, hence it has the same eigenvalues. So if $DB$ is symmetric, it is positive definite.</p> <p>I note, however, that a diagonally dominant matrix is not necessarily positive definite, although it has eigenvalues of positive real part.</p>

http://mathoverflow.net/questions/47603/other-ways-to-define-naturals/47608#47608 Michael Renardy 2010-11-28T20:10:11Z 2010-11-28T20:10:11Z

<p>It depends what "express in terms of" means. Are the following allowed?</p> <p>$$S(x)=x+f_1(f_4(x)),$$ $$Pd(x)=x-f_1(f_4(x)).$$ Or perhaps something like: $$S(x)=f_2(f_4(x)+f_1(f_4(x))).$$</p>

http://mathoverflow.net/questions/47314/bounding-a-smooth-function-near-the-endpoint/47321#47321 Michael Renardy 2010-11-25T11:51:09Z 2010-11-25T11:51:09Z

<p>You need to make the stronger assumption $g(a)=g'(a)=...=g^{(k-1)}(a)=0$. Then your statement is true with $\alpha=k$. You can see this by using the Cauchy-Schwarz inequality in $g^{(k-1)}(x)=\int_a^x g^{(k)}(y)\,dy$ to obtain $|g^{(k-1)}(x)|\le C(x-a)^{1/2}$, and then integrating repeatedly to get $|g(x)|\le C(x-a)^{k-1/2}$.</p> <p>This is essentially optimal, since the function $(x-a)^{k-1/2}/\ln(x-a)$ satisfies all the hypotheses. In particular, you cannot get $\alpha=k+1$. </p>

http://mathoverflow.net/questions/46104/a-simple-ordinary-differential-equation/46114#46114 Michael Renardy 2010-11-15T13:00:32Z 2010-11-15T13:00:32Z

<p>You don't need the Cauchy-Kovalevskaya theorem. Just the analytic inverse function theorem.</p>

http://mathoverflow.net/questions/50472/sums-of-arctangents Michael Renardy 2010-12-27T21:21:38Z 2010-12-27T21:21:38Z

No, the OEIS stuff does not pan out. The modulus of the next coefficient is 3, not 5.

http://mathoverflow.net/questions/50472/sums-of-arctangents Michael Renardy 2010-12-27T11:51:16Z 2010-12-27T11:51:16Z

You can reexpand the Taylor series of the arctan function. I am not sure what the pattern is. Up to tenth order, I get $$\arctan(x)=\arctan(1)+\arctan((x-1)/2)-\arctan((x-1)^2/4)+\arctan((x-1)^3/8)$$ $$-\arctan((x-1)^5/32+\arctan((x-1)^6/64)-\arctan((x-1)^7/128$$ $$+\arctan((x-1)^9/256)-\arctan(3(x-1)^{10}/1024).$$ So up to this point, we get the coefficient sequence $$1, -1, 1, 0, -1, 1, -1, 0, 2, -3.$$

http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product/50125#50125 Michael Renardy 2010-12-22T04:02:50Z 2010-12-22T04:02:50Z

The problem as I understood it did not say B was symmetric, only that H was.

http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product Michael Renardy 2010-12-22T03:52:58Z 2010-12-22T03:52:58Z

BD and DB are similar matrices, so they have the same eigenvalues.

http://mathoverflow.net/questions/47418/nice-classes-of-non-closable-operators Michael Renardy 2010-11-26T14:18:53Z 2010-11-26T14:18:53Z

Moreover, a closed operator can have empty spectrum.

http://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals/47021#47021 Michael Renardy 2010-11-23T00:40:09Z 2010-11-23T00:40:09Z

All you need to do is prove that between two rationals is an irrational. A variant of the well known proof that sqrt(2) is irrational should do the trick here. Just exploit the sparsity of squares among &quot;large&quot; integers.