Laplacian's Posts
Nairaland Forum › Laplacian's Profile › Laplacian's Posts
1 2 3 4 5 6 7 8 9 (of 9 pages)
 Education › 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+  75=4475since 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 |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 4:27pm On Jan 01, 2015*. Modified: 4:45pm 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. |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 8:21pm On Nov 06, 2014*. Modified: 9:21pm On Nov 06, 2014 |
mnairaland:this can be solved either geometrically or algebraically. By geometry, it is easier since u are required to find the coordinate of intersection of two circles. Here's an algebraic proof; d12=(x-x1)2+(y-y1)2.......eqn1 d12=(x-x1+x2-x2)2+(y-y1+y2-y2)2 d12=(x-x2+(x2-x1))2+(y-y2+(y2-y1))2 or d12=(x-x2)2+2(x-x2)(x2-x1)+(x2-x1)2+(y-y2)2+2(y-y2)(y2-y1)+(y2-y1)2 substitute equation (2) in this last eqn and call it eqn (3), equation (3) will involve (x-x2) and (y-y2). From equation (3), substitute for (y-y2) in eqn (2) and solve the resulting quadratic in (x-x2) bros, the alphabet involved is too much, so dont expect the final solution here |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 7:46am On Nov 05, 2014 |
Contd.
|
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 7:42am On Nov 05, 2014 |
Here's question 1.)
|
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 7:28am On Nov 05, 2014 |
emmyeuler1:
|
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 11:38am On Oct 03, 2014 |
jackpot: Dear Math Generals, Lieutenants, Colonels, and friends of Maths, please help out:sum of the radii must be greater than the distance between centres!!! 1 Like |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 11:14am On Oct 03, 2014 |
akpos4uall: O we maths enthusiast, I'm sure some of us are already familiar with some software/applications that can be used in solving problems. I'm familiar with MATLAB only. From the little I've learnt, I must say it'll be nice if we all can learn how to make use of it or any other similar application.tanks |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 8:17am On Jul 31, 2014 |
Freiburger: https://www.nairaland.com/1834634/need-mathematics-questions1.) Given 3x-2x=65, we use Modular arithmetic, working in Mod3 we get; -2x # -1 (Mod3) Or (-1)x # 1 (Mod3) where # means "congruent to". The congruence holds iff x is even. Hence, x=2k, and we have 32k-22k=65 Or (3k-2k)(3k+2k)=65*1=13*5. Now there are four possibilities; (3k+2k)=65 (3k-2k)=1 Or (3k+2k)=1 (3k-2k)=65 Or (3k+2k)=5 (3k-2k)=13 Or (3k+2k)=13 (3k-2k)=5 the first three eqns have no solution, adding or subtractin d last eqn gives k=2 and hence x=2k=2*2=4 |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 3:19pm On Jul 30, 2014 |
benbuks: (x/2)^x = 16given; (x/2)x=16=24, by the unique factoriztaion theorem, (x/2) must be a power of two since the right hand side of the given eqn is a power of two. Hence, (x/2)=2k for some integer k. Hence, 2kx=24 or kx=4 and there are only three possibilities; k=1, x=4 Or k=x=2 Or k=4, x=1. Testing each option in the given equation shows that only the first one is valid, Hence x=4 |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 9:51pm On Jul 17, 2014 |
jackpot: Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of triangle ABC. Derive the formula for the coordinates of it's circumcentre.general equation of a circle, x2+y2-2ax-2by+c=0, the centre is (a,b), three points on the circumference are given. This is enough!! |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 6:20pm On Jul 11, 2014 |
benbuks: Y^x+x^y=17 ..........(i)if u are interested in only positive integer solutions (it is an easy matter to show that negative integers cannot satisfy both equations), then i can give a simple solution |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 3:29pm On Jan 05, 2014 |
...can the product of two distinct, non-zero integers when increased by one, divide the sum of their squares?...solutions&suggestions 're welkom.... |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 1:37pm On Jan 05, 2014*. Modified: 3:30pm On Jan 05, 2014 |
djkay1000: a.bros dis one na 4 astrophysicists ooo...*hands on chin, sighin, waitn patiently 4 som1 2 unravel d mystery* 1 Like |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 7:19am On Jan 05, 2014*. Modified: 7:39am On Jan 05, 2014 |
Mr Calculus: NO 3 HAS NO SOLUTION CAUSE R1 WHICH IS IN D QUES IS NOT IN D VALUE 4 Lpermit me 2 use i for j (R1+iwL)/R3=R2/(R4-i/wC) or (R1+iwL)/R3=R2*(R4+i/wC)/((R4)2+(1/wC)2) or (R1+iwL)=R3*R2*(R4+i/wC)/((R4)2+(1/wC)2) but Mr. Calculus had already shown that if z, a, b, c, and d are Real and z=(a+ib)/(c+id), then a/b=c/d now set a=R1, b=wL, c=R4, d=1/wC and z=R3*R2/((R4)^2 + (1/wC)2) then all the conditions of questn 1 are satisfied. Hence, R1/wL=R4*wC, so L=R1/(R4*w2C). IF U DOUBT THE RESULT, SUBSTITUTE REAL NUMBERS TO VERIFY |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 6:53pm On Jan 04, 2014*. Modified: 7:23pm On Jan 04, 2014 |
Pat £inst£in:i thought u 've resolved it: now, f(x,z)=f(x)+zf'(x)+z2f"(x)/2!+... If f(x,z)=tan-1(x+z), then the above equation resolves into tan-1(x+z)=tan-1x+zf'(x)+z2f"(x)/2!+... Since f(x)=tan-1x then, f'(x)=1/(1+x2) f"(x)=-2x/(1+x2)2=-2x[f('x)]2 f'"(x)=d(-2x[f('x)]2)/dx, e.t.c now since cot(z)=x, => cot2(z)=x2, or 1+x2=1+cot2z=cosec2x, :. sin2x=1/(1+x2), i hop u can complet the rest *i must comment that, your series is slightly INCORRECT* |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 5:22pm On Jan 01, 2014 |
Pat £inst£in:k, we know that, for a series to be representable in a binomial form, the terms must have a recurrent structure, and the numerator & denominator must have the same number of factors. The two conditions are satisfied for case u solved, since Tn=#(2r+1)/[4n-1(n-1)!]=Tn-1*(2n+1)/[4n-1(n-1)!] the one i solved also to satisfies: Tn=#(2r+3)/[3n+1*(n+1)!]=Tn-1(2n+3)/[3n+1*(n+1)!], but it fails for the nunber of factors, to accommodate that, consider the eqn; (1+x)n=1+nx+[n(n-1)/2!]x2+[n(n-1)(n-2)/3!]x3+... I think a little modification can make it suit the conditions e.g (1+x)n-(1+nx)=[n(n-1)/2!]x2+[n(n-1)(n-2)/3!]x3+... We divide thru by nx to obtain [(1+x)n-(1+nx)]/nx=[(n-1)/2!]x+[(n-1)(n-2)/3!]x2+[(n-1)(n-2)(n-3)/4!]x3+... Use ur method for the last eqn lets c hw far it goes... |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 3:52pm On Jan 01, 2014 |
Pat £inst£in:yea, i made the error while typin, u postd ur comment while i was editin mine, its already correctd. nice one! So u 're in possession of d solution afterall |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 1:47pm On Jan 01, 2014*. Modified: 3:12pm On Jan 01, 2014 |
Pat £inst£in:for ur no.1, am not really sure what u mean by "odd and even terms", but here's a suggestion; 1.) expansion of (x+a)b should contain only "odd & even terms", so that, from the condition of the question, a+b=(x+a)b, then multiplying thru by a-b, we obtain; a2-b2=(x+a)b•(a-b).. 2.) this one is quiet 'cheap', given f(x)= (1 - x + x2)4. Take the fouth order derivative of the above expression, set x=0, and divide the result by 4!... Simply put, obtain fiv(0)/4! In case u ain't satisfied with that, here is an elementary ALTERNATIVE SOLUTION (1 - x + x2)4=[1+(x-1)x]4=1+4(x-1)x+6[(x-1)x]2+4[(x-1)x]3+[(x-1)x]4, inspection shows that only the last three terms contains x4, and the coefficient can be readily obtained from there. 3.) let # denote pi, the product of an expression in d variable r, from r=1 to n, by inspection, the numerator of each term of ur series is the product of an A.P with fomula; #(3+2r), similar, the denominator is the product of the natural numbers and some power of three, with the formula; 3n+1•(n+1)!, so the nth term of ur sequence is; Tn=[#(3+2r)]/[3n+1•(n+1)!] Recall that, #(3+2r)=5•7•9•11•13•15... =#[(2+2r)•(3+2r)]/#(2+2r) =(4•5•6•7....)/(4•6•8•10•12...) #(3+2r)= [(3+2n)!/3!]/(2n•3•4•5•6...) =[(3+2n)!/3!]/(2n/2•[(3n+2)/2]!) so that, Tn=[(3+2n)!/3!]/(2n/2•[(3+2n)/2]!)/[3n+1•(n+1)!] the solution becomes too complicatd to paste here afterward...i'll paste the final solution soon... |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 9:45am On Jan 01, 2014 |
benbuks: ^ ok...m on bed nw..mayb i ‘d try dem 2maro...quit cheep sha.if u tink they are cheap, den we 'll patiently wait 4 ur solution... |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 7:26am On Dec 18, 2013 |
akpos4uall: I noticed the mistake while I was fixing in the values....yea, i 've corrected that, u posted ur comment just as i was about to modify it... Ur result is very impressive...pls i need complete details (very elaborate) on how u generated those programs on ur PC...are u on Facebook?... |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 10:21pm On Dec 17, 2013*. Modified: 12:09am On Dec 18, 2013 |
[quote author=akpos4uall]I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him. 1, 8, 15 4, 30, 56 15, 112, 127 56, 418, 780 2911, 21728, 40545 Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples. This can be written as X, 2Y, Z Y, 2Z, 4Z - 2Y Z, 8Z - 4Y, 15Z - 8Y 4Z - 2Y, 30Z - 16Y, 56Z - 30Y 15Z - 8Y, 112Z - 60Y, 209Z - 112Y 56Z - 30Y, 418Z -224Y, 209Z - 112Y,[/quote]i luv ur observation..nice wrk!...if u get d proof, then u hav found another algorithm 4 Pell's eqn |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 8:54am On Dec 17, 2013*. Modified: 11:06am On Dec 17, 2013 |
Laplacian: For those who may be confused, what he has written mayb replicatd as follows;proofs: PROPOSITION 1: proof: let xy be a two digit HAPPY number, then the followin sequenc holds; x2+y2=abcd a2+b2+c2+d2=mnpqsr ..... ..... ..... =10000....0 Suppose we have xy*10k for some integer k, then xy*10k=xy00000...0 with the zeros in k places, hence x2+y2+02+02+...+02=abcd, and the above sequenc is repeatd and the result follows. COROLLARY 1: there are infinitely many HAPPY numbers. Proof; because k is a natural number and there are infinitely many natural numbers, the result follows PROPOSITION 2: proof: let xyz be HAPPY, then x2+y2+z2=abcd, where abcd is HAPPY, now the permutation yxz is shows that y2+x2+z2=x2+y2+z2=abcd, {because addition is cummutative}, and the result follows.. COROLLARY 2: if any number of zeros is either struck out or fixed in between the digits of a HAPPY number, the new number formed is also HAPPY. Proof: case 1: let the number be xyz0000...0 havin k zeros, then by the above proposition, the number can be permuted to take m zeros to the left of the natural numbers, i.e 000...0xyz000..0=xyz000...0, where the zeros to the right of xyz is k-m, for k>m. e.g, 2300 is HAPPY (by proposition 1), so 0230=230 is also HAPPY. case 2: if xyz is HAPPY, then by proposition 1, xyz*10k =xyz000...0 is HAPPY, so permutin we have, x000...0y00..0z000...0 PROPOSITION 3: proof: obvious PROPOSITION 4: proof; the number 111...000, with m ones and n zeros follows the undelisted sequenc; 12+12+12+...+02+02+02=1+1+1+...+0+0+0=m, hence if m is HAPPY, the number is also happy COROLLARY 3: there are infinitely many binary numbers proof: from above, for any HAPPY number m, there exist a binary number containin m ones, or equivalently, there exist a one to one correspondenc between HAPPY numbers and binary numbers, but from COROLLARY 1 above, there are infinitely many HAPPY numbers, and the result follows. PROPOSITION 5: proof: obviously follows from 3.) PROPOSITION 6: proof: let xy and ab be two numbers such that x+a<=9, y+b<=9 , suppose both numbers are HAPPY, then x2+y2=pqrs, is HAPPY and a2+b2=uvwz is also HAPPY. Now xy+ab=(x+a)(y+b) so the sequenc of HAPPINESS for the sum is; (x+a)2+(y+b)2=x2+a2+2ax+y2+b2+2by=x2+y2+a2+b2+2(ax+by), now the number 2(ax+by) is not alwaz HAPPY (why?, because the product of HAPPY or SAD numbers is not alwaz happy, by proposition 3 above), and the result follows. In particular, if ab and xy are HAPPY then ab+xy is HAPPY iff 2(ax+by) is HAPPY. The same reasonin goes for a subtraction. PROPOSITIO 7: Loading... Anyone who proves the last gets a reward from me...promise... Hey!, my proofs above are open 2 criticism 1 Like |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 5:45pm On Dec 16, 2013*. Modified: 11:03am On Dec 17, 2013 |
For those who may be confused, what he has written mayb replicatd as follows; Alpha Maximus: For example, 19 is happy, as the associated sequence is:My little observations, proofs to follow shortly; 1.) if a number is HAPPY, then, that number multiplied by any integral power of ten is also HAPPY. e.g 7 and 70 and 700 2.) if a number is HAPPY, the number formed by permutation of the digits of the original number is also HAPPY. e.g 19 and 91 or 49 and 94 3.) the product of any two HAPPY numbers may/may not be HAPPY. e.g 7*19=133 4.) a binary number is HAPPY iff the nimber of 1 it contains is itself HAPPY 5.) the set of ALL HAPPY numbers is not closed under the multiplicative binary operation, and so is not a Group. 6.) the sum and difference of any two HAPPY number is not a HAPPY number 7.) if two numbers are HAPPY, then the number formed by combinin the digits of the two original numbers is not HAPPY |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 12:58pm On Dec 16, 2013 |
echibuzor: Before Jackpot gives that, can you also provide the steps u used in getting your results?i support u sir |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 11:34pm On Dec 15, 2013 |
Alpha Maximus: Confirmed! But can you explain how you arrived at that formula?since d sequence is too scanty to admit a rule she approximtd let xn=a+bn+cn2 substitut 4 each of x1, x3 and x6 she obtained three equtns in which she solved simultaneously for a, b, and c...& i agree with her |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 1:10pm On Dec 15, 2013 |
Humphrey77: SIMPLIFY (9-0)(9-1)(9-3)...pls simplify it lets c @humphrey |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 11:47am On Dec 15, 2013 |
jackpot: Clear.hey, i have more unanswered problems than any user on this thread , few in number theory though (as compared 2 elliptic functions)...here are two interestin ones, i've cracked d first one though, maybe u can generalise my result lik u did 4 alpha maximus, or mayb ur method of solution 'll get 2 b prettier than mine, in any case i'll get 2 learn smthin....subsequently, in all my questions, whenever i refer 2 integers, i mean NON-ZERO integers to avoid the supply of trivial solutions: 1.) find the set of all integer pairs (x, y) such that x2+y2 is divisible by xy+1 2.) if 4n+2n+1 is a prime, show that n is some power of 3 |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 7:16pm On Dec 14, 2013 |
Laplacian: @jackpot, 2nioshine&Dx....tanks 4 sharin my burden...believin everyone understnds Dx's solutn, i proceed 2 give my solutn cuz i went farther & wait 4 ur correctns...for any confusion, refer 2 page 121 1 Like |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 6:18pm On Dec 14, 2013*. Modified: 7:20pm On Dec 14, 2013 |
jackpot: Hahaha. I didnt know your first statement (the part where you replied Maximus) is still on the question. I thought you were boilingi'll post the full solution later (wen am free). I arrived @ it exactly 12:00pm today...it has 2 do wit Pell's equation...i hop u 're familiar wit that?...hey! the error has been duely corrected...thanks... |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 5:26pm On Dec 14, 2013 |
jackpot: This one no be proof na....i 'll assume this statement is not comin 4rm jackpot...i suggest the fool who has jackpot's phone should return it 2 her, because jackpot will not make such a futile comment,,, |
| Education › Re: Nairaland Mathematics Clinic by Laplacian(m): 4:36pm On Dec 14, 2013*. Modified: 7:22pm On Dec 14, 2013 |
Alpha Maximus: Well, what's it then?for any natural number n, the expression (2-√3)n can be expressed in the form d-k√3....now d is the required common difference, the required first term, a=d-2k... @jackpot...if there are infinite natural numbers, then there are infinite solutions |
