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Re: Nairaland Mathematics Clinic by Nobody: 6:55am On Jun 09, 2015 |
benji93:Thanks for pointing that out. Corrected. |
Re: Nairaland Mathematics Clinic by Tolzeal(m): 10:27am On Jun 09, 2015 |
guys, Please what do think it cool as elective course?? Complex Analysis or abstract Algebra. Please candid opinion will be appreciated . |
Re: Nairaland Mathematics Clinic by Nobody: 12:51pm On Jun 09, 2015 |
Tolzeal:They are both interesting. You should choose the one you find easier. If you do want to experience a whole new world in mathematics though, then choose abstract algebra. If you are a physics major and need extra tools for your physics, then complex analysis is the way to go. Goodluck. |
Re: Nairaland Mathematics Clinic by Tolzeal(m): 1:07pm On Jun 09, 2015 |
goofyone: Thaanka... |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 2:25pm On Jun 09, 2015 |
The gurus are resurfacing ... nice work guys..., wish I have enough time...will try sha.. BTW: Benkuks=agentofchange1 @ sir richiez I sight thee boss. 1 Like |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 2:26pm On Jun 09, 2015 |
The gurus are resurfacing ... nice work guys..., wish I have enough time...will try sha.. BTW: Benkuks=agentofchange1 @ sir richiez I sight thee boss. |
Re: Nairaland Mathematics Clinic by naturalwaves: 6:35pm On Jun 09, 2015 |
agentofchange1:Sup Benbuks? |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:29pm On Jun 09, 2015 |
naturalwaves:Am cool. 1 Like |
Re: Nairaland Mathematics Clinic by frob0genius(f): 9:37pm On Jun 09, 2015 |
Tolzeal: To add to what goofyone wrote. Abstract Algebra has more of theorems, lemma and proves. It deals with algebraic structures rather than the usual number systems we all know, unlike complex analysis. |
Re: Nairaland Mathematics Clinic by Soneh(m): 10:14pm On Jun 11, 2015 |
goofyone:please boss check the division ,i have been trying to follow up the solution but couldn't come up with same answer as ur's don't knw if i'm doing the wrong thing |
Re: Nairaland Mathematics Clinic by naturalwaves: 10:30pm On Jun 11, 2015 |
Soneh:Let me help you.... 1 Like
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Re: Nairaland Mathematics Clinic by Soneh(m): 10:47pm On Jun 11, 2015 |
naturalwaves:thanks .if you don't mind can u pls help with the explanation to this 3. Prove that the equation mx(x2 + 2x + 3) = x2 – 2x – 3 has exactly one real root if m = 1 and exactly 3 real roots if m= -2/3 For this question, you can use Descartes's rule of signs. You should actually use the more detailed Sturm's extension of this rule, but i think the former could be okay here. Descartes's rule of signs says if v is the variation of sign in a polynomial p(x) and the number of positive real zeros (roots) is n, then n<=v and v-n is even. Same applies to number of negative real zeros with p(-x) With m=1, you get x3 + x2 + 5x + 3 = 0 p(x) = x3 + x2 + 5x + 3 No sign variation, so v=0. n will also be 0 so there are no positive real roots. p(-x) = -x3 + x2 - 5x + 3 Number of sign variation is 3 (- to + to - to +) n will be 1, so that 3-1 is even. So there's a negative real root. Equation thus has one negative real root and two complex roots which will be conjugates. With m= -2/3 p(x) = -(2/3)x3 - (7/3)x2 + 3 Number of sign variation is 1. n will be 1 so that 1-1=0 which is even. Hence, one positive real root. p(-x) = (2/3)x3 - (7/3)x2 + 3. Number of sign variation here is 2. n cannot be 1, cos v-n should be even. So n is 2. And 2-2=0. So we have two negative real roots. Hence, three real roots (one positive and two negative) |
Re: Nairaland Mathematics Clinic by ikoyila: 12:55pm On Jun 12, 2015 |
I FELT LIKE CRYING WHEN I SAW THIS FORUM: MATHEMARICS. MY BEST SUBJECT. AM SO FOLLOWING, IF WORK WILL ALLOW ME. 1 Like |
Re: Nairaland Mathematics Clinic by naturalwaves: 3:24pm On Jun 12, 2015 |
Soneh:To be honest, I do not know much about those principles but I could read through his solution and try and break it down for you in my spare time. |
Re: Nairaland Mathematics Clinic by donriddo(m): 10:13pm On Jun 12, 2015 |
Wow! I cant believe where i just landed... Nice meeting you all Sirs/Mas. I look forward to learning more than a lot from you. Please help me. 6Px = 30. Where P = permutation |
Re: Nairaland Mathematics Clinic by naturalwaves: 10:59pm On Jun 12, 2015 |
donriddo:The answer is 2. See solution attached below...... 2 Likes
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Re: Nairaland Mathematics Clinic by dejt4u(m): 11:03pm On Jun 12, 2015 |
donriddo: 6Px = 30, 6! / (6 - x)! = 30, 30 x 4! = 30(6 - x)!, (6 - x)! = 4!, Obviously, x = 2 Q.E.D 2 Likes |
Re: Nairaland Mathematics Clinic by donriddo(m): 11:52am On Jun 13, 2015 |
dejt4u: naturalwaves:Thanks my ßosses. I really appreciate. 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 1:04pm On Jun 13, 2015 |
naturalwaves:Good solution, naturalwaves. But I don't think you can mathematically cancel "!" signs like that. It's just sufficient to say (6-x)=4, I believe. |
Re: Nairaland Mathematics Clinic by DatechMan(m): 1:16pm On Jun 13, 2015 |
Can someone figure out an analytical solution to this equation? 4^x + 3^x = 5^x |
Re: Nairaland Mathematics Clinic by Seriallinks(m): 4:49pm On Jun 13, 2015 |
^^^ The answer is 2. But I can only give you graphical solution! |
Re: Nairaland Mathematics Clinic by Nobody: 5:29pm On Jun 13, 2015 |
DatechMan:It's much easier to prove that x cannot be less than 2, and x cannot also be greater than 2. Hence x = 2 for all real numbers. |
Re: Nairaland Mathematics Clinic by killsmith(f): 5:59pm On Jun 13, 2015 |
1) prove that every metric space is a hausdorff space.....2) show that the empty intersection of sets (the finite intersection of empty sets) is the universal set.... |
Re: Nairaland Mathematics Clinic by naturalwaves: 6:09pm On Jun 13, 2015 |
goofyone:thanks sir, what you have up there and mine are the same just that you cancelled yours and didn't indicate while I showed mine. We both cancelled it. I thought about it too and came up with the conclusion that it can be cancelled provided that the l.h.s equates to the r.h.s. For example, if x! = y!, then x must be = y. If x! = (y+z)!, then x must be equal to (y+z) and so on and that was why I cancelled it. I wouldn't mind any example that disproves these facts. Thanks. 1 Like |
Re: Nairaland Mathematics Clinic by donriddo(m): 7:04pm On Jun 13, 2015 |
DatechMan:4^x + 3^x = 5^x divide bothsides by 4^x 1 + (3/4)^x = (5/4)^x 1 + 0.75^x = 1.25^x 1 + (1-0.25)^x = (1+0.25)^x Approximating linearly 1 + 1 - 0.25x = 1 + 0.25x 2 - 0.25x = 1 + 0.25x 0.5x = 1 x = 2 1 Like 3 Shares |
Re: Nairaland Mathematics Clinic by naturalwaves: 7:40pm On Jun 13, 2015 |
@goofyone, please help Soneh with the explanation of that Descarte's principle and how you used it to solve that problem,he doesn't seem to understand it. Thanks |
Re: Nairaland Mathematics Clinic by Nobody: 8:11pm On Jun 13, 2015 |
naturalwaves:Yeah, i understand you. It's okay, bro. 1 Like |
Re: Nairaland Mathematics Clinic by dejt4u(m): 5:49am On Jun 14, 2015 |
Thank God you guys are back to level I cn comprehend.. The fear of madam Jackpot's questions is d beginning of learning..lol 1 Like |
Re: Nairaland Mathematics Clinic by Nwiboazubuike(m): 8:00am On Jun 14, 2015 |
Integrate cos^4x dx |
Re: Nairaland Mathematics Clinic by frob0genius(f): 4:01pm On Jun 14, 2015 |
Nwiboazubuike: Made a mistake in the last line it's sin 2x over 4 not over 8
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Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:18pm On Jun 14, 2015 |
Nwiboazubuike:^^^^^^^^SolutioN^^^^^^^^^^ let's rewrite as $(cos2 x )2dx known from trig. ( Pythagorean) identities , cos2 =0.5(cos2x+1) thus ,=> $[(0.5(cos2x+1)]^2 dx => 1/4$((cos^2 (2x) +2cos2x +1))dx we can also have cos^2(2x) = 0.5(cos4x +1) thus , get have 1/4$[0.5(cos4x +1) +2cos2x +1] dx now on integrating we get => 1/4[ 1/2*1/4*sin4x + x/2 +sin2x + x ] +k => 1/32* sin4x +1/4*sin2x+ 3x/8 + k |
Re: Nairaland Mathematics Clinic by STENON(f): 7:23am On Jun 16, 2015 |
naturalwaves:thanks, STENON is in the mood of her final exams. Good morning, Hope you are good? 1 Like |
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