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Mathematics Clinic / 2015 Cowbell Mathematics Champion, Akinkuowo Honoured By School. / Lead City University Clinic Welcomes First Ever Baby Since 10 Years Of Opening (2) (3) (4)
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Re: Nairaland Mathematics Clinic by Nobody: 7:11am On Jan 01, 2015 |
Miscellaneous: Yes sire! |
Re: Nairaland Mathematics Clinic by jackpot(f): 10:14am On Jan 01, 2015 |
doubleDx:2nd order, yes. Nonlinear, yes. But I doubt if it is of degree 1. To see this, recall that sin(y'')=y''- (y'')3/3!+ (y'')5/5!-(y'')7/7!+. . . So, inherently, we see that the degree is infinite (or does not exist?) that's what I feel ooooh. Please, criticize. ![]() Happy New Year, Sire. Happy New Year, NMCites ![]() 2 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 12:30pm On Jan 01, 2015 |
jackpot: You are actually right! The degree doesn't exist... I have rechecked! Thanks for pointing that out... Happy new year! It's been a while....howdy? |
Re: Nairaland Mathematics Clinic by Laplacian(m): 4:27pm On Jan 01, 2015 |
jackpot: It Is a NON-LINEAR, SECOND DEGREE, THIRD ORDER differential equation PROOF; given; sin(y'')+y'=0 or sin(y'') = -y'............(1.) first we differentiate the above using function of a function to get; d[sin(y" )]/dy" x dy"/dt + d(y')/dt=0 where we take t as the independent variable; we obtain cos(y" ) x y'" + y"=0 or cos(y" ) = -y"/y'" ..............(2.) square (1.) and (2.) and add; sin2(y'' ) + cos2(y'' )= (y')2 + (y"/y'" )2=1 or [1-(y')2] x (y'" )2-(y" )2=0 Q.E.D if u doubt that the two differential equations are inequivalent, solve the first one (since it is easier) and show that it satisfies the second. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 5:37pm On Jan 01, 2015 |
A RULE ON MULTIPLICATION suppose we are to multiply two numbers with 5 as unit digit each. We proceed as follows; let the numbers be n5 & m5 then n5 x m5 = (10n+5) x (10m+5) = 100nm+50(n+m)+25=100nm+100(n+m)/2+25=100[nm+(n+m)/2]+25 Hence; n5 x m5 = [nm+(n+m)/2]25 Theorem; i.) to multiply two numbers ending with 5, take the product of their non-unit digit, take the average of their non unit digit, sum them and add 25 @ the end. Provided the two non-unit digits are either both odd or both even. E.g 135 x 15= (13+7)25=2025 since 13 and 1 are both odd, 13 x 1 =13 and (13+1)/2=7 so 13+7=20 again, 25 x 65= (12+4)25=1625 since 2 and 6 are both even, 6 x 2 =12 and (6+2)/2=4 so 12+4=16 ii.) if one of the non-unit digit is odd and the other is even, take their product, add the non-units to 1 and divide by 2, sum the result with the product, put 75 @ the end. E.g 125 x 35= (36+ ![]() since 12 and 3 have opposite parity, 12x3=36, and (12+3+1)/2=8 COLORALLY; i.) if two or more numbers ending with 5 are multplied, the last two digits of the result must either be 25 or 75 ii.) to multiply a number by 5, if its non unit digit is even, divide it by 2 and add 25. E.g 65 x 5=(6/2)25=325. It its non-unit digit is odd, add 1 to it and divide by 2 and put 75 behind the result. E.g 35x5=(3+1)/2=275. All these follow from the above result; n5xm5=[nm+(n+m)/2]25 if m=0 then n5x05=[0+(n+0)/2]25=(n/2)25 iii.) to square a number that ends with 5, multply the non-unit digit by the nearest integer greater than it. From our formula; n5xm5=[nm+(n+m)/2]25 put n=m then, m5xm5=(m5)2=[m2+m]25=m(m+1)25 e.g 652=(6x7)25=4225 or 852=(8x9)25=7225 or 152=(1x2)25=225 or 52=052=(0x1)25=025=25 or 352=(3x4)25=1225 or e.t.c 2 Likes |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 5:58pm On Jan 01, 2015 |
Laplacian: hay man , the prof. himself. . but seems its of a non-existence /infinite degree since even after every successive derivatives we obtain different D.E ......we employ your approach for P.S ( power series solutions) . hope you get the gist ..? you're great man !! |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:02pm On Jan 01, 2015 |
greetings ..... happy new year please help out with this. integral of ln(sinx) dx = ? |
Re: Nairaland Mathematics Clinic by 2nioshine(m): 6:13pm On Jan 01, 2015 |
Laplacian:Highly creative! |
Re: Nairaland Mathematics Clinic by Laplacian(m): 12:20am On Jan 02, 2015 |
agentofchange1:hi ben! Let I=§In(sinz)dz, let y=In(sinz).........(1.), dy/dz=cosz/sinz, so , I=§ytanz.dy............(2.) from (1.) ey=sinz or e-2y=cosec2z so that (e-2y-1)1/2=cotz so that (2.) becomes; I=§y/(e-2y-1)1/2dy or I=§ yey/(1-e2y)1/2dy if we integrate by part by keepin y constant first we get; I=-y(1-e2y)1/2+§(1-e2y)1/2dy now let H=§(1-e2y)1/2dy let u=ey so that du/dy=u Hence H=§(1-u2)1/2du/u Or H=§(u-2-1)1/2du now let u=cosw then du/dw=-sinw, Hence, H=§tanw.-sinwdw or H=§-sin2w/cosw dw =§(cos2w-1)/cosw dw =§(cosw - 1/cosw) dw =sinw-§1/cosw dw let G=§1/cosw dw =§cosw/cos2w dw let k=sinw then dk/dw=cosw hence G=§1/(1-k2) dk using partial fraction and resolvin we get G=1/2 In[(1+k)/(1-k)] = 1/2 In[(1+sinw)/(1-sinw)] therefore H=sinw-1/2 In[(1+sinw)/(1-sinw)] but I=-y(1-e2y)1/2+H hence I=-y(1-e2y)1/2+sinw-1/2 In[(1+sinw)/(1-sinw)]+Constant where w=cos-1(ey) and y=In(sinz) simplifying further if u wish, I=[1-In(sinz)]cosz +1/2 In[(1+cosz)/(1-cosz)]+constant |
Re: Nairaland Mathematics Clinic by jackpot(f): 4:38pm On Jan 02, 2015 |
Laplacian:hmmm. I can see the mathematics. But. . . Let's take another look. Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation y''+y'=0. It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e-x, whereas y''+y'=0 has the solution y=C1+C2e-x. Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution. So, I will say that the two differential equations you compared are not equivalent. Or, what do you think? |
Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 4:56pm On Jan 02, 2015 |
jackpot: ![]() |
Re: Nairaland Mathematics Clinic by mathefaro(m): 5:38pm On Jan 02, 2015 |
jackpot:Wait o, are you sure you're actually feminine? |
Re: Nairaland Mathematics Clinic by Laplacian(m): 8:12pm On Jan 02, 2015 |
jackpot: y should they not be equivalent? Integration as we know is d inverse of differentiation. Does it not imply that if one differentiates a function and integrates it he should still have the same general solution? Let's use ur example. Given y'+y=0 after differentiating, y"+y'=0, should we not have same function (in a more general form) if we integrate? §y"dt+§y'dt=§0xdt so that y'+y=C1 and this is slightly more general and does no harm to the original function (in terms of generality). Let's solve the two equations; from dy/dt=C1-y we get dy/(C1-y)=dt so that -In(C1-y)=t+K or C1-y=e-t-K or y=C1-e-Kxe-t so that y=C1+C2e-t. This tells u how C1 came into the picture. And as i have explained above, the are one and the same. Questioning its validity amounts to doubtin if the following is true; given y=t2+3........(1.) differentiate, y'=2t, now let's integrat d last to get y=t2+C............(2.) and u say (1.) and (2.) are not the same? Leave it or take it INITIAL VALUE SETTLES EVERYTHING MY DEAR. So on what ground do u accuse me of heresy, and what is the FOUNDATION of ur claim of inequivalence? |
Re: Nairaland Mathematics Clinic by jackpot(f): 9:23pm On Jan 02, 2015 |
Laplacian:Both differential equations won't even have the same number of initial conditions. The one you differentiated will need extra one initial condition. For example, y'+y=0 needs only one initial condition before you obtain the particular solution free from arbitrary constants, whereas y''+y'=0 needs two initial conditions. So, i am thinking it makes them even far from being equivalent? 1 Like |
Re: Nairaland Mathematics Clinic by Laplacian(m): 9:47pm On Jan 02, 2015 |
jackpot:C1 came as a result of an integration (see above). So u are saying in essence that, differentiating a function and integratin the result will not yield the initial function. Because i differentiate; y'+y=0 to get y"+y'=0 and i integrate this to get y'+y=C1 |
Re: Nairaland Mathematics Clinic by jackpot(f): 12:23am On Jan 03, 2015 |
Laplacian:maybe in the world of cartoons(where things happen the way we want), we may say that y'+y=C is equivalent to y'+y=0 for all values of the constant C. ![]() ![]() 1 Like |
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:10am On Jan 03, 2015 |
jackpot:everyone is entitled to his opinion |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:43am On Jan 03, 2015 |
Laplacian:hmmm. Suppose for contradiction that y'+y=0 is equivalent to y''+y'=0. By differentiating once more, we must have that y''+y'=0 is also equivalent to y'''+y''=0. But these guys are an equivalence relation, so by transitivity, y'+y=0 is equivalent to y''+y'=0 which is equivalent to y'''+y''=0 implies that y'+y=0 is equivalent to y'''+y''=0. Now, y'+y=0 has the solution y=C-x, but y'''+y''=0 has the solution y=C1+C2x+C3e-x. Are these two solutions equivalent? ![]() 1 Like |
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:46am On Jan 03, 2015 |
jackpot:yes! A simple substitution C1=C2=0 will pay the price. It stil boils down to initial value. |
Re: Nairaland Mathematics Clinic by jackpot(f): 7:12am On Jan 03, 2015 |
Laplacian:using your logic, y=Ce-x is equivalent to y=C1ex+C2e-x which is (again, by your logic) equivalent to y=Cex. Then, by transitivity of an equivalence relation, y=Ce-x is equivalent to y=Cex. Do you really believe that both solutions are equivalent? ![]() |
Re: Nairaland Mathematics Clinic by Laplacian(m): 7:40am On Jan 03, 2015 |
jackpot:what extension of the D.E for y=Ce-x did u get the solution y=C1ex+C2e-x if u want to play by my logic? |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:03am On Jan 03, 2015 |
hmmm. think am enjoying this. drama. as long it doesn't leads to something else . hmm ride profs . na wa ooo. Differential equations sha .. ^^hay guys those are. mathematical models that assume a real life phenomenon , why not use both model to solve a problem let's see what happens .? just a suggestion sha .. ## just passing by c.ya lera . 1luv |
Re: Nairaland Mathematics Clinic by jackpot(f): 8:07am On Jan 03, 2015 |
Laplacian:since y'+y=0 is equivalent (you said so) to y''+y'=0, it is equivalent to y''+y'-(y'+y)=0, which is the same thing as y''-y=0. The DE y''-y=0 now has the solution y=C1ex+C2e-x. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 9:05am On Jan 03, 2015 |
jackpot:but the solution of y'+y=0 always satisfies any of its extended D.E doesn't it? But the matter of the extended D.E's solution (y"-y=0) satisfying the original D.E (either y'+y=0 or y'-y=0) is decided by initial value. The values on ex are not on e-x, so with d initial value inputed in the solution of the extended D.E, either C1 or C2=0 as appropriate. Turning tables around, i ask u; if the D.E y'+sin(y" )=0 is of infinite degree, why should the D.E obtained from it have degree 2? |
Re: Nairaland Mathematics Clinic by jackpot(f): 12:46pm On Jan 03, 2015 |
Laplacian:There's no theorem saying that if you differentiate a DE, the degree will increase or decrease. In other words, there is no theorem comparing the degree of a DE with that of the differentiated DE. So, you shouldn't compare both degrees. Maybe you're trying to draw analogy from the fact that the difference between the order of a DE and with that of the differentiated DE is 1. Catch my drift, Sir? 1 Like |
Re: Nairaland Mathematics Clinic by Laplacian(m): 5:47pm On Jan 03, 2015 |
jackpot:i catch ur drift ma. But, with due respect, just because u've not seen a theorem on that does not erase d fact that it does exist. And do i always need a theorem to confirm what i know? |
Re: Nairaland Mathematics Clinic by tohero(m): 7:06pm On Jan 03, 2015 |
Battle of the Titans! ![]() They made me thought if I ever sat in a mathematics class. |
Re: Nairaland Mathematics Clinic by Emmafe(m): 1:58pm On Jan 04, 2015 |
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Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:04pm On Jan 04, 2015 |
Evaluate |5 1 0 0 0| |1 4 0 1 2| |3 2 1 4 0| |0 1 2 1 0| |1 2 0 1 2| |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:06pm On Jan 04, 2015 |
tohero: no be small tin oo |
Re: Nairaland Mathematics Clinic by Nastydroid(m): 8:34pm On Jan 04, 2015 |
tohero:no be only you |
Re: Nairaland Mathematics Clinic by tohero(m): 9:38pm On Jan 04, 2015 |
Nastydroid: agentofchange1:All is well. At least we sabi count 1, 2, 3... So we arnt getting freaked! |
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