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Re: Nairaland Mathematics Clinic by MaxGraviton: 5:05am On Aug 03, 2015 |
Geofavor: I never said a computation of 5C0 was invalid. I was only pointing out that accounting for that which was not present was incorrect. (Of course I've been proved wrong). I now see my mistake. Geofavor: And I'm still making it clear that, if you want to prove to anyone that 0 has value, FACTORIAL of 0, is not a way. Whether in combination & permutation or in Kangaroo Maths, (lools) . 5C0 = 5!/(5 - 0)! 0![/quote] |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:09am On Aug 03, 2015 |
possibilita: ok here Q1c y' = (3y2 +2xy)/(2xy+x2 ).. ......(B) Set y=vx. V= y/x => y'=xv' +v ...(E) Equate with (B) above We have Xv'+v= [3(vx)^2 +2x(vx)]/[2x(vx) +x^2] =( 3x^2 v^2 +2x^2 v )/(2x^2 v +x^2) Pull out ' x^2 ' => xv' =(3v^2 +2v) /(2v+1). - v = xv '= (3v^2 +2v -2v^2 -v)/(2v+1) Xdv/dx = (v^2 +v) /(2v +1) =>( 2v+1)dv/(v^2 +v) = dx/x Taking integral on both sides yields In(v^2 +v)=lnx + lnC Or v^2 +v = CX Since y/x = v hence we have =(y/x)^2 + (y/x) = CX ii) y' = y/x + xsin(y/x) Let vx=y , v= y/x Xv'+ V = y' We get xv' +v = v + xsinv . Or v'= sinv By separating variables => (1/sinv)dv = dx On integrating both sides => In[tan(v/2)] = x+c Or tan(v/2)= exp(x+c) Or v/2 =arctan[exp(c+x)] Or v= 2arctan[ e^c *e^x] Since v=y/x =>y/x = 2arctan(Ke^x) (Where k= e^c) Hence Ans:y=2xarctan(Ke^x) That's it man. 3 Likes 1 Share |
Re: Nairaland Mathematics Clinic by possibilita(m): 7:17am On Aug 03, 2015 |
agentofchange1:thank u very much bros |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:24am On Aug 03, 2015 |
possibilita: UWC man 1 Share
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Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:34am On Aug 03, 2015 |
m kinda busy now, remaining solutions might come later , av solved them already , just 2uplaod ... or my bosses in d hus can do that , #Slalom. |
Re: Nairaland Mathematics Clinic by Antoinne: 8:54am On Aug 03, 2015 |
agentofchange1:Good job, agentofchange1. |
Re: Nairaland Mathematics Clinic by Antoinne: 9:16am On Aug 03, 2015 |
agentofchange1:There seems to be a mistake in the first two lines in the first picture with y2. Would affect the final answer significantly. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:21am On Aug 03, 2015 |
Antoinne: oh yeah, my bad , just noticed that now ...will upload corrected solutions sir thanks once again. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 12:16pm On Aug 03, 2015 |
jackpot: considering the denominator s^5 +a^5 in general xn + yn =(x+y) £ yk* x(n-1-k) for k=0,1 2, . . . ,(n-1) where £ = sigma (summation) does that ring a bell?. I know say you na Prof. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 12:28pm On Aug 03, 2015 |
Antoinne:yes, i think they will b equal. This new diagram luks beta |
Re: Nairaland Mathematics Clinic by MaxGraviton: 12:35pm On Aug 03, 2015 |
Please, This is really Urgent. Someone should please help me with the name of that Maths textbook popularly known as BONDE. . . Bonde is the name of the person that wrote the book. But Please I need the actual name. Thanks. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 12:38pm On Aug 03, 2015 |
In spherical co-ordinate, with standard notation, is it not logical to expect d elementary area on d sphere to be; dS=r2d@d¤? Why is it necessary to resolve in the xy plane to have; dS=r2sin@d@d¤. Note; @ is theta, ¤ is phi |
Re: Nairaland Mathematics Clinic by Antoinne: 5:01pm On Aug 03, 2015 |
MaxGraviton:Pure Mathematics for Advanced Level by B.D. Bunday & H. Mulholland |
Re: Nairaland Mathematics Clinic by Antoinne: 5:04pm On Aug 03, 2015 |
Laplacian:Come on! No, they will not be equal. Look at the diagram again. "A" will have to be A.cos(phi) = dx. Not A = dx. You see it? |
Re: Nairaland Mathematics Clinic by MaxGraviton: 5:47pm On Aug 03, 2015 |
Antoinne: May God Bless Your Generation. . . thanks. |
Re: Nairaland Mathematics Clinic by jackpot(f): 10:52pm On Aug 03, 2015 |
agentofchange1:**in a kid's voice** Broda, finish it |
Re: Nairaland Mathematics Clinic by Laplacian(m): 5:52am On Aug 04, 2015 |
Antoinne:u'r right, they form a right angled triangle...maybe i shoul just take a rest! |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:01am On Aug 04, 2015 |
please, help me out. 1. Consider the parabola: x=(3-2y)(4y+3). Obtain i. Directrix equation ii. Focus iii. Vertex iv. Length of it's latus-rectum 2. Consider the hyperbola 4x^2-4x-y^2+6y+28=0. Obtain i. Eccentricity ii. Foci iii. Vertex iv. Centre v. The Length of it's semi-latus rectum vi. It's asymptotes. tags: agentofchange1, dejt4u, Laplacian, Antoinne, Karmanaut, doubleDx, etc |
Re: Nairaland Mathematics Clinic by Antoinne: 6:17am On Aug 04, 2015 |
Laplacian:I like your inquiring mind though. Naturally i wouldn't have done it the way you intended to do it. Most textbooks wouldn't, since it's just easy to use the dot product. But your approach to understanding it from "first principle" is commendable. Still give it a try when you free. Just think about resolving the elemental area and not the individual distances. Let me see if there's an approach to resolving the individual distances and I'll get back. 1 Like |
Re: Nairaland Mathematics Clinic by jackpot(f): 4:03pm On Aug 04, 2015 |
Please help me out Solve the functional equations below for f(x) and g(x) respectively: 1. f(x)-f(-x)= - 1/x 2. g(x)-g(-x)=-1/x, given that g is odd. tags: agentofchange1, dejt4u, STENON, AmazingAngel, Antoinne, Laplacian, Karmanaut, doubleDx, etc |
Re: Nairaland Mathematics Clinic by Laplacian(m): 4:22pm On Aug 04, 2015 |
jackpot:let kn=e2#ni/5, n=1,...,5 where # is pi then, 1/(s^5+a^5)=product1/(s-akn)....resolve using partial fraction, cover-up rule (all factors are distinct) e.t.c |
Re: Nairaland Mathematics Clinic by Nigeriatraining: 4:44pm On Aug 04, 2015 |
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Re: Nairaland Mathematics Clinic by Laplacian(m): 5:06pm On Aug 04, 2015 |
jackpot:if f(x)-f(-x)=-1/x then f(x) can neither be an even function nor have an even part, otherwise we have f(-x)=f(x) or f(x)-f(-x)=0, plug into d given eqn we have 0=-1/x, takin inverse makes x undefined. If f(x) is odd then f(-x)=-f(x), hence 2f(x)=f(x)-f(-x)=-1/x |
Re: Nairaland Mathematics Clinic by Youngsage: 5:08pm On Aug 04, 2015 |
O boy! |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:01pm On Aug 04, 2015 |
Laplacian:Now, this is what I call a sexy solution. Introduce i but make the solution look compact. Unlike the other solution that is complex upon complex with all the known operations (imaginary, plus, minus, division, multiplication, square root, so many lines , etc) in mathematics coming in a single solution. Well done Sir 1 Like |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:11pm On Aug 04, 2015 |
Laplacian:Well, unlike integers that are either even or odd, a function may be neither even nor odd, eg h(x)=sin x-cos x, x+x2. Also, functions may have non-integral exponent, eg h(x)= xe=x2.718281828. . . or p(x)= £ x, where £ is the Gamma function or q(x)=x2/7. Now, my question is: how are we sure that there are no neither-even-nor-odd functions f satisfying the equation: f(x)-f(-x)= -1/x ? tags: Laplacian, agentofchange1, Laplacian , agentofchange1, doubleDx, Antoinne, dejt4u, AlphaMaximus, Laplacian etc |
Re: Nairaland Mathematics Clinic by Antoinne: 6:58pm On Aug 04, 2015 |
jackpot: Parabola of x = (3-2y)(4y+3) x = 12y + 9 - 8y2 - 6y x = 9 + 6y - 8y2 Parabola is about y, since y has two values. We can treat this as a mirror y = 9 + 6x - 8x2, and then invert the mirror at the solution, if x is easier to work with. anyways, for a parabola y = ax2 + bx + c with a focus at p distance above the vertex V(h,k). dy/dx = 2ax + b. At the vertex, you have the least variation, so dy/dx=0 2ax + b = 0 x = -b/2a, this should correspond to h. y = ax2+bx+c at x = -b/2a, k k = (4ac - b2)/4a Assuming parabola at origin and following definition 4py = x2 Move parabola to right of xy plane so that vertex is V(h,k) 4p(y-k) = (x-h)2 Relating coefficients produces p = 1/4a, (focus) Latus-rectum is length across focus. if Parabola at origin (0,0) 4py = x2 with y = p on curve x2 = 4p.p = 4p2 x = 2p Length of latus-rectum 2.x = 4p Now, from earlier equation y = 9 + 6x - 8x2 (mirror) a = -8, b = 6, c=9 focus, p = 1/4a = 1/(4*-8 ) =-1/32 Vertex, V(h,k) h = -b/2a = -6/(2*-8 ) = 3/8 k = (4ac - b2)/4a = 81 / 8 Length of latus-rectum = 4p =abs(4.(-1/32)) = 1 / 8 Directrix equation: y = 325/32, i.e. (1/32 + 81/8 ) We can now go ahead and mirror our solution by just changing all x to y, and y to x i. Directrix equation: x = 325/32 ii. Focus: -1/32 iii. Vertex: V(81/8, 3/8 ) iv. Length of it's latus-rectum: 1/8 1 Like |
Re: Nairaland Mathematics Clinic by Laplacian(m): 7:18pm On Aug 04, 2015 |
jackpot:# with functions that are neither even nor odd, one can often find each part seperately and add them together. # functions in which the variables are in integral powers are often called algebraic functions, and the theory of functional equations is usually restricted to this class of functions...elliptic functions, theta functions, mathieu functions e.t.c are not elementary functions and exhibit very cmplx properties which are unfamiliar and when incorperated into this subject makes it irrelevant and uninteresting # there is no claim as to the uniqueness of a solution to a functional equation, since there maybe other solutions depending on the available properties and method of solution used |
Re: Nairaland Mathematics Clinic by Antoinne: 7:25pm On Aug 04, 2015 |
jackpot:I think you right. We cant know unless otherwise stated. And I think you provided a solution already if f(x) were odd. 1 Like |
Re: Nairaland Mathematics Clinic by Laplacian(m): 7:38pm On Aug 04, 2015 |
Antoinne:Assuming parabola at origin and following definition y = 4px2? |
Re: Nairaland Mathematics Clinic by Antoinne: 7:50pm On Aug 04, 2015 |
Laplacian:Corrected. thanks |
Re: Nairaland Mathematics Clinic by Nobody: 7:54pm On Aug 04, 2015 |
jackpot: |
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