Hi,
Does someone have idea how to derive from :
to 2*sqrt(E/5) -(y-3*sqrt(E/5))^2
1/sqrt(Pi*No) * Integral [e^ ------------------ ]
from -oo No
to :
to +oo
1/sqrt(Pi) * Integral e^(-z^2) dz
from sqrt(E/(5*No))
Can we write (y-3*sqrt(E/5))^2)
------------------ = z ??? If so what then?
No
Thanks and best regards
Integral derivation ...
Started by ●March 24, 2009
Reply by ●March 24, 20092009-03-24
On Tue, 24 Mar 2009 11:31:09 -0500, Melinda wrote:> Hi, > > Does someone have idea how to derive from : > > to 2*sqrt(E/5) -(y-3*sqrt(E/5))^2 > 1/sqrt(Pi*No) * Integral [e^ ------------------ ] > from -oo No > > to : > > to +oo > 1/sqrt(Pi) * Integral e^(-z^2) dz > from sqrt(E/(5*No)) > > > Can we write (y-3*sqrt(E/5))^2) > ------------------ = z ??? If so what then? > No > > Thanks and best regardsThen we open the calculus book to the section on "variable substitution" (sometimes called "u substitution") and we brush up on the basics. -- http://www.wescottdesign.com
Reply by ●March 24, 20092009-03-24
On Mar 24, 12:31�pm, "Melinda" <melinda.m...@gmail.com> wrote:> Hi, > > Does someone have idea how to derive from : > > � � � � � � � � to 2*sqrt(E/5) � � �-(y-3*sqrt(E/5))^2 > 1/sqrt(Pi*No) * Integral � � �[e^ � ------------------ � ] > � � � � � � � � from -oo � � � � � � � � � �No > > to : > > � � � � � � �to +oo > 1/sqrt(Pi) * Integral � � � � � �e^(-z^2) dz > � � � � � � �from sqrt(E/(5*No)) > > Can we write �(y-3*sqrt(E/5))^2) > � � � � � � � ------------------ = z �??? � If so what then? > � � � � � � � � � � � �No > > Thanks and best regardsThank goodness that in the real world you don't have to deal in closed form integration. Just plug the formulas in a computer and get the answer. I think closed form integration is one of those things all engineers need to learn to toss in the garbage can. It is really used as a weapon by mathematicians to belittle us. But in reality, practically nothing solves out in clsed form. Learn the ideas behind the integration, visualize the math and never look back at your inability to do a closed form integration problem ever again.
Reply by ●March 24, 20092009-03-24
On Mar 24, 9:29�pm, spamkiller1...@gmail.com wrote:> On Mar 24, 12:31�pm, "Melinda" <melinda.m...@gmail.com> wrote: > > > > > > > Hi, > > > Does someone have idea how to derive from : > > > � � � � � � � � to 2*sqrt(E/5) � � �-(y-3*sqrt(E/5))^2 > > 1/sqrt(Pi*No) * Integral � � �[e^ � ------------------ � ] > > � � � � � � � � from -oo � � � � � � � � � �No > > > to : > > > � � � � � � �to +oo > > 1/sqrt(Pi) * Integral � � � � � �e^(-z^2) dz > > � � � � � � �from sqrt(E/(5*No)) > > > Can we write �(y-3*sqrt(E/5))^2) > > � � � � � � � ------------------ = z �??? � If so what then? > > � � � � � � � � � � � �No > > > Thanks and best regards > > Thank goodness that in the real world you don't have to deal in closed > form integration. > > Just plug the formulas in a computer and get the answer. > > I think closed form integration is one of those things all engineers > need to learn to toss in the garbage can. �It is really used as a > weapon by mathematicians to belittle us. �But in reality, practically > nothing solves out in clsed form. > > Learn the ideas behind the integration, visualize the math and never > look back at your inability to do a closed form integration problem > ever again.- Hide quoted text - > > - Show quoted text -Your world is different from mine. Often people bring real world engineering problems to me to solve for money. So go ahead and tell people they don't need to learn the math. I don't mind! Besides the integral the OP asked about is simple to solve. Just square it and write as a product of two integrals. Then use Fubini's theorem to write that as a double integral. Next change over to polar coordinates from cartesian. It then becomes an elementary integral that can be done by simple u substitution. Then finally find the square root of the result. This particular integral shows up in many areas of engineering and physics. The learning of these techniques is not to belittle anyone, but rather to enable whole classes of problems to be solved. How do you think the numerical methods get checked? Clay
Reply by ●March 25, 20092009-03-25
clay@claysturner.com wrote: (after someone wrote)>> Thank goodness that in the real world you don't >> have to deal in closed form integration.(snip)> Your world is different from mine. Often people bring real world > engineering problems to me to solve for money. So go ahead and tell > people they don't need to learn the math. I don't mind!I agree. Even if we don't have to use it so often, it is still useful to know, at least the simple integrals. (Or even how to look them up in an integral table.) Numerical evaluation of integrals give approximate answers, and uses up a lot of computer time. If a closed form solution is available, then use it! -- glen
Reply by ●March 25, 20092009-03-25
On Mar 24, 10:58�pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:> Numerical evaluation of integrals give approximate > answers, and uses up a lot of computer time. �If a closed > form solution is available, then use it!In this instance, a closed-form solution is not possible except as an infinite series which would then have to be evaluated numerically for given value of E/N0. The integral in question is essentially (i.e. except possibly for a missing constant) the value of the cumulative probability distribution function (CDF) of a Gaussian random variable. This CDF has been proven (yes, by those pesky mathematicians that spamkiller complains about) to be *not* expressible as a finite formula involving the elementary functions (exp, log, sin, cos etc). Those same mathematicians have also come up with easy-to-compute approximations for this CDF, and one of these approximations is built into most "scientific" calculators on the market. Finally, the OP's problem can be solved by a simple change of variables as Tim Westcott suggested earlier. --Dilip Sarwate
Reply by ●March 25, 20092009-03-25
On 25 Mar, 13:29, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:> In this instance, a closed-form solution is not possible > except as an infinite series ...> Finally, the OP's problem can be solved by a simple > change of variables ...Seems you are contradicting yourself here? Rune
Reply by ●March 25, 20092009-03-25
Hi and thanks, So guys how we change those integration limits from -inf to 2*sqrt(E/5), to --> from sqrt(E/(5*No)) to +inf. Someone said from cartesian to polar coordinates? But this is not like Pythagoras theorem and angles stuff, because it is from -inf not some point to point. Thanks
Reply by ●March 25, 20092009-03-25
On Mar 25, 10:57�am, Rune Allnor <all...@tele.ntnu.no> wrote:> On 25 Mar, 13:29, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote: > > > In this instance, a closed-form solution is not possible > > except as an infinite series ... > > Finally, the OP's problem can be solved by a simple > > change of variables ... > > Seems you are contradicting yourself here? > > RuneOK, folks, from the top..... Given two expressions F(x) and G(x), does F(x) equal G(x)? For example, does x^2 + 2x + 1 equal 4x? The answer is No as mathematicians know, but engineers who grab their calculators and check whether equality holds, say at x = 1, might be misled. A persistent and cautious engineer may try x = 2 next and discover the error, but time, as many posters in this thread have pointed out, is of the essence, and we want to get "the answer" as soon as possible and with as little effort as possible.... Does the integral from -infinity to b of f(x) equal the integral from h(b) to +infinity of g(x)? Just evaluating the two integrals numerically for a specific value of the parameter b might give the wrong answer because equality (or maybe equality to the numerical accuracy of the integration routine) happens to hold for that choice of the parameter b but not for other choices of b. In the instance asked about by the OP, the numerical value of b is not specified, and so the question is really of the type whether functions F(b) and G(b) are equal. Now, both integrals can be expressed as infinite series in b. Are the two series the same? Once again, we could evaluate the series for some chosen value of b, but could be misled. But, **for the specific problem asked for by the OP**, the correct answer can be easily determined *****without finding the numerical value of either integral**** by using the change of variable method suggested by Tim Westcott, one of the earliest responders in this thread. The integral from -infinity to b of (1/s)exp[-{(x-m)^2}/{2s^2}] equals, by a change of variables y = (x-m)/s, the integral from -infinity to (b-m)/s of exp[-{y^2}/2] , and by a further change of variables z = - y, the integral from (m-b)/s of exp[-{z^2}/2]. It is to be hoped that the OP will be able to figure out the answer to her question from this additional information and put this thread to rest. Sheesh! --Dilip Sarwate
Reply by ●March 25, 20092009-03-25
dvsarwate@yahoo.com <dvsarwate@gmail.com> wrote: (I wrote)>> Numerical evaluation of integrals give approximate >> answers, and uses up a lot of computer time. ?If a closed >> form solution is available, then use it!> In this instance, a closed-form solution is not possible > except as an infinite series which would then have to > be evaluated numerically for given value of E/N0.Yes. I was commenting on the general idea of not needing closed form solutions. Many useful integrals don't have closed form solutions, but you wouldn't know that if you weren't interested in finding a closed form solution at all. -- glen
