tag:blogger.com,1999:blog-5159574672638793285.post431022985767782655..comments2024-01-29T06:01:15.858-05:00Comments on The Center of Math Blog: Advanced Knowledge Problem of the Week Center of Mathematicshttp://www.blogger.com/profile/06584574038886207370noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-5159574672638793285.post-48335157725018952852016-11-18T08:03:08.241-05:002016-11-18T08:03:08.241-05:00Thanks for the equations. These were quite helpful...Thanks for the equations. These were quite helpful.<br /><a href="https://www.mobitairportparking.co.uk/car-parking-manchester" rel="nofollow">secure airport parking Manchester</a><br /><a href="https://www.mobitairportparking.co.uk/car-parking-manchester" rel="nofollow">Manchester parking deals</a>Suzan Bakerhttps://www.blogger.com/profile/08076528232999266758noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-22174771868262633752016-06-09T14:56:25.970-04:002016-06-09T14:56:25.970-04:00I was looking for the solution and It was a great ...I was looking for the solution and It was a great help. <br /><a href="https://www.easymeetandgreetluton.co.uk/" rel="nofollow">parking luton airport</a> Easy Meet and Greet Lutonhttps://www.blogger.com/profile/14690951161172803888noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-27242320045293777392016-06-09T07:01:52.429-04:002016-06-09T07:01:52.429-04:00These were the most annoying problem sums for me e...These were the most annoying problem sums for me ever. <br /><a href="http://www.bestmeetandgreetgatwick.co.uk/" rel="nofollow">gatwick chauffeur parking</a> Anonymoushttps://www.blogger.com/profile/17282895038220897807noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-62112194924125883862016-06-06T04:25:05.600-04:002016-06-06T04:25:05.600-04:00This comment has been removed by a blog administrator.Heathrow Terminal 4 Parkinghttp://www.bestmeetandgreetheathrow.co.uk/terminalsnoreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-88972949055826925242016-04-18T07:28:13.119-04:002016-04-18T07:28:13.119-04:00This comment has been removed by a blog administrator.Anonymoushttps://www.blogger.com/profile/17878744154041890320noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-38581811331635876362015-12-31T01:36:52.805-05:002015-12-31T01:36:52.805-05:00I also split the integral into -0.5 * (integral r ...I also split the integral into -0.5 * (integral r * -2*r*e^(-r^2)) and solved that by partial integration to -0.5 * (r*e^(-r^2) - integral e^(-r^2)) with boundaries of 0 to infinity. By knowing that the integral e^(-r^2) from 0 to infinity is sqrt(pi)/2 and by solving r*e^(-r^2) for infinity to be 0 and r*e^(-r^2) for 0 to be 0 too, it simplifies to -0.5 * (0 - (-sqrt(pi)/2)), which is sqrt(pi)/4sgruenwaldhttps://www.blogger.com/profile/11564906004927744406noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-14153783764459483832015-12-02T13:48:37.778-05:002015-12-02T13:48:37.778-05:00The integral is essentially the second moment of a...The integral is essentially the second moment of a Gaussian random variable with mu =0 and sigma^2 = 1/2. Then the result is immediate, since we have Second moment = sigma^2 = 2/sqrt(pi) {problem integral}Sigmahttps://www.blogger.com/profile/05070806742281193958noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-61573059372899670432015-12-02T13:38:16.257-05:002015-12-02T13:38:16.257-05:00When solving standard Gaussians I always try to av...When solving standard Gaussians I always try to avoid iterated integrals. IBP is quite a bit faster. Also, and i didn't work it out all the way, but another trick is to square the non-standard Gaussian to get the same result, again only needing to evaluate one integral. <br /><br />Just some tricks from the physics side of things. Jasonnoreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-65357713178861846452015-12-02T12:46:41.951-05:002015-12-02T12:46:41.951-05:00Good point, Judith! There's often more than on...Good point, Judith! There's often more than one way to solve any problem, and sometimes I can miss what might seem obvious to you. Center of Mathematicshttps://www.blogger.com/profile/06584574038886207370noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-66346352400969769502015-12-02T09:28:45.172-05:002015-12-02T09:28:45.172-05:00I used partial integration, with f=r and g'=r*...I used partial integration, with f=r and g'=r*e^(-r^2). Seems way more easy to me?Judithhttps://www.blogger.com/profile/00934536740335557689noreply@blogger.com