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Life lessons from differential equations (2015)
- garmaine 8y ago> Some problems simply have no solution Don't confuse "exact solution" with "solution" (unqualified). There is no problem that has no solution. I stopped reading there.
- lixtra 8y agoDepends on what you call a solution. If you call there is no such function a solution, then yes every well posed problem has a solution. But many people call only an object that exists a solution and has no solution just means no such object exist. If you take ill posed problems - the barber that shaves everyone that doesn’t shave themselves. Then you are no longer able to answer with true/false so in the limited context there is indeed no solution.
- jfoutz 8y agoseven bridges?
- gaius 8y agoGo around the source of the river It’s not a good solution, but it is one nonetheless!
- prmph 8y agoWhat is the solution for avoiding death and taxes?
- severino 8y agoDying is not always a problem. In fact, lots of people consider it a solution. The same can be said about taxes. They were the solution for problems like giving people access to education and healthcare.
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- Irinopolis 8y agoBecoming that which deals out death and taxes?
- jk4930 8y agoFollow Peter Thiel into the future: https://www.newyorker.com/magazine/2011/11/28/no-death-no-taxes https://www.newyorker.com/magazine/2011/11/28/no-death-no-ta...
- csomar 8y agoI'm not certain about death but lots of people are not paying taxes. Don't let others convince you otherwise.
- ginnungagap 8y agoPlenty of problems have no solution, even in PDEs land where we're quite willing to redefine what's a solution (weak solutions, distributions etc.)
- paulpauper 8y agoboundary conditions do indeed make solving differential equations much harder thankfully the use of infinite series can help a lot in such situations
- ianai 8y agoIt’s important to try many different approaches to difficult problems. Or, try different problems with new approaches - to learn about the approaches.
- chrbarrol 8y agoI felt that one addition could be "Many problems have trivial solutions, but they are of little practical value" 0 is often a trivial solution to many PDEs, but it is of little analytical value, so it is often discarded. In the same sense death is a trivial of many life problems: Dying would solve most (all?) problems in life, but it is not a solution you would typically consider :)
- stabbles 8y agoIt all depends on the boundary conditions.
- goldenkey 8y agoMy boundary condition is fulfilling the categorical imperative of pure energy. For some people its the imperative of themselves or immediate families, or mammals or life itself. I still haven't figured out what the universe would like but I am at its service. My best guess at the current moment is to extend the universe's life but the inevitable heat death if correct throws a wrench in that. However if there is a way to create continuity from discreteness, and we can simulate our universe, we may be able to run a child universe to completion before our own universe dies. And if the same thing happens in that child, we will have had an infinite number of universes live and die in the finite life of our own universe. If that isn't a full life for a universe, I don't know what is... This already may be happening if black holes birth and contain universes.
- smaddox 8y ago> I still haven't figured out what the universe would like but I am at its service. If you're trying to figure out "what the universe would like", you're going to be searching until the heat death of the universe. > However if there is a way to create continuity from discreteness, and we can simulate our universe, we may be able to run a child universe to completion before our own universe dies. This is ontologically unworkable. The sooner you accept that nothing will last forever, the sooner you can get to living your life and enjoying the things that are here right now.
- Yhippa 8y agoOne of the most important lessons I learned was spending several pages of notebook paper devising a clever solution only to find that it completely doesn't work. Pretty darn good way to humble yourself.
- esturk 8y agoOne of the points link to a second article that reads: (The fact that φ is zero outside a finite interval mean the “uv” term from integration by parts is zero.) Can anyone elaborate on this? I'm not too sure how this trick works.
- pedrosorio 8y agoThis is the article you're referring to: https://www.johndcook.com/blog/2009/10/25/how-to-differentiate-a-non-differentiable-function/ https://www.johndcook.com/blog/2009/10/25/how-to-differentia... This is the wikipedia page on integration by parts: https://en.wikipedia.org/wiki/Integration_by_parts https://en.wikipedia.org/wiki/Integration_by_parts He's setting u = phi(x), v = f(x). Since he's integrating on the real line, a = -infinity, b = +infinity, so on the right hand side for "uv" you'd have u(+inf)v(+inf) - u(-inf)v(-inf) - and since u (which is phi) is 0 outside of a finite interval, you know these terms are 0. [You obviously can't evaluate functions at +-inf, and you have to take limits to evaluate u(x)v(x) for an improper integral, but you can see the result is the same if u(x) is zero outside of a finite interval]
- esturk 8y agoOh I see. So in the evaluation of u(+inf), it becomes 0, hence uv is 0. I have never thought of it that way. Thanks!