Nairaland Mathematics Clinic - Education (172) - Nairaland
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| Re: Nairaland Mathematics Clinic by Videx1: 11:07pm On Apr 11, 2015 |
agentofchange is that all about moment generating function |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 11:18pm On Apr 11, 2015 |
Videx1:sir, you want me to post a text book on it here right ? hmm., c'mmon bruv , surf the net na .? or bring up questions on it . guess that would help . |
| Re: Nairaland Mathematics Clinic by jackpot(f): 12:55pm On Apr 12, 2015 |
agentofchange1:i use to tell the junior ones that it is easier to destroy a house than to build. Same way it's easier to differentiate than to integrate. |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 2:18pm On Apr 12, 2015 |
jackpot:yea , exactly the point. |
| Re: Nairaland Mathematics Clinic by Cymbal: 4:20pm On Apr 12, 2015 |
Kindly help me with this. ..."Prove that the equation R=COS@ represent equation of a circle"... TAGS: agentofchange1 ,jackpot,benbuk,laplacian,dejt4u, and evry math guru. THANKS. |
| Re: Nairaland Mathematics Clinic by Cymbal: 5:51pm On Apr 12, 2015 |
Find in what ratio the points; (4b-2a , 9c-a) divides the line joining the points (a+b , 3c+5a) and (5b-3a , 11c-3a). |
| Re: Nairaland Mathematics Clinic by jackpot(f): 6:15pm On Apr 12, 2015 |
Cymbal: Cymbal: Cymbal:r= cos@= x/r implies r2=x implies r2-x=0 implies x2-x+y2=0 implies (x- 1/2)2+(y-0)2= (1/2)2 This is an equation of circle centred at (1/2, 0) with radius 1/2. |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:29pm On Apr 12, 2015 |
How will I go about this plz..?. The function f(x,y) satisfies 1) f(0,y)=y+1 2)f(x+1,0)=f(x,1) 3)f(x+1,y+1)=f(x,f(x+1,y)) for all non negative integers x,y. determine f(4,1981) |
| Re: Nairaland Mathematics Clinic by jackpot(f): 8:33pm On Apr 12, 2015 |
agentofchange1:google functional equations |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:47pm On Apr 12, 2015 |
jackpot:try this dear let's see. |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:27pm On Apr 12, 2015 |
Cymbal:3:1 |
| Re: Nairaland Mathematics Clinic by Cymbal: 9:42pm On Apr 12, 2015 |
agentofchange1:kindly show the workings! |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:49pm On Apr 12, 2015 |
Cymbal:x=(pX2+qX1)/(p+q) y=(pY2+qY1)/(p+q) Where x,y={(4b-2a),(9c-a)} X1,Y1={(a+b),(3c+5a)} X2,Y2={(5b-3a),(11c-3a)} 4b-2a={(5b-3a)p+(a+b)q}/(p+q) P(a-b)=3q(a-b) P=3q hence p:q=3:1 |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 9:53pm On Apr 12, 2015 |
Nature130:xy+yx=17 and x+y=5, using modular arithmetic, and letting # denote "congruent to" we have; xy # 17 (mod y) and x # 5 (mod y) hence; 5y # 17 (mod y). Since y is relatively prime to 5 and 17, we have, by Fermat's Little Theorem, 5y # 5 # 17 (mod y), hence, 12 # 0 (mod y), so y is a positive divisor of 12 and is less than 5, the only candidate are y= 2 or 3 or 4, same argument goes for x since the equation is symetrical. So, x= 2, or 3, or 4 and y=2, or 3, or 4. Suppose x<y, then, 2x<x+y=5 hence x<2.5. If x is less than x then x=2 and hence, from y=5-x, y=3 |
| Re: Nairaland Mathematics Clinic by Cymbal: 9:56pm On Apr 12, 2015 |
jackpot:thanks a bunch.. |
| Re: Nairaland Mathematics Clinic by jackpot(f): 10:02pm On Apr 12, 2015 |
jackpot: |
| Re: Nairaland Mathematics Clinic by jackpot(f): 10:04pm On Apr 12, 2015 |
Laplacian:another angle of attack, that's cool |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 10:14pm On Apr 12, 2015 |
agentofchange1:x4=4x first; (x4)1/x=4, since the RHS is an integer, x must divide 4; similarly; x=(4x)1/4, again since LHS is an integer, 4 must divide x. Hence, if x divides 4 and 4 divide x, then x=4. For; 3x=8x, since 3 cannot divide 8, no integer x can satisfy that equation |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:14pm On Apr 12, 2015 |
jackpot:d guy like modular arithmetic too much .hhmm @ prof. Laplacian I hail oo. |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:21pm On Apr 12, 2015 |
Laplacian:yea man , got that , but you see the 2nd 1 becomes a challenge , that's why I said there exist no general rule to solving such . |
| Re: Nairaland Mathematics Clinic by Cymbal: 10:45pm On Apr 12, 2015 |
agentofchange1:thanks!!! |
| Re: Nairaland Mathematics Clinic by Cymbal: 10:50pm On Apr 12, 2015 |
Help me with this : Find the equation of the circle which touches the axis of y at a distance +4 from the origin and cuts off an intercept 6 from the axis of x. TAGS: agentofchange1 ,jackpot,laplacian. |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 11:14pm On Apr 12, 2015 |
Nature130:4x=8x or 22x-3=x this clearly shows that x must be some power of 2, i.e x=2t for some positive integer t. Hence, 22^(t+1)-3=2t, or 2t+1-3=t, now t cannot divide 2 otherwise it divides 3, so using Fermat's Little Theorem, we have; 2t+1-3 # 0 (mod t) or 2*2t-3 # 0 (mod t) or 2*2-3 # 0 (mod t) or 4-3 # 0 (mod t) or 1 # 0 (mod t) hence, t is a positive divisor of 1. Hence, t=1, but x=2t. Hence, x=2 |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 11:54pm On Apr 12, 2015 |
Cymbal:at least three points are required to specify a circle, so we must determine the second intercept on the x-axis; if the second intercept is k, then; k=42/6=8/3 so the three points are; (0,4), (6,0), (8/3,0). Hope it helps!! |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 12:13am On Apr 13, 2015 |
[quote author=jackpot post=32654806][/quote]hi jackpot. Question 3 is analogous to a ball rolled off a cliff with initial velocity equal to that of the space craft |
| Re: Nairaland Mathematics Clinic by Cymbal: 12:29am On Apr 13, 2015 |
Laplacian:it helps buh please kindly give me full solution and explain how k=8/3.THANKS. |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 1:01am On Apr 13, 2015*. Modified: 7:16am On Apr 13, 2015 |
agentofchange1:startin wit the third, set x=0 then; f(1,y+1)=f(0,f(1,y)), now apply rule 2 to the RHS to get f(1,y+1)=1+f(1,y). And this is all we need, we can apply this recurrence y times to get; f(1,y+1)=1+f(1,y)=2+f(1,y-1)=...=y+1+f(1,0). Hence, f(1,y)=y+f(1,0)............eqn4 from eqn2, put x=0, then; f(1,0)=f(0,1). from rule1, f(0,1)=1+1=2. Hence, f(1,y)=y+2..............eqn5 so eqn1 and eqn5 give us; f(0,y)=y+1 and f(1,y)=y+2. Next, set x=1 in eqn3 and use eqn5. Repeat the process until u get to f(4,y) and set y=1981 to get your result. I will not solve it completely cos i don't have the time. |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 1:04am On Apr 13, 2015 |
Cymbal:heard of Newton's theorem for the radius of curvature of a line? |
| Re: Nairaland Mathematics Clinic by Cymbal: 1:09am On Apr 13, 2015 |
Laplacian:nah |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 1:14am On Apr 13, 2015 |
Cymbal:ok. Just google "intersecting chord theorem" because i can't draw the diagram here |
| Re: Nairaland Mathematics Clinic by Cymbal: 1:15am On Apr 13, 2015 |
Laplacian:please kindly complete the question afta getting the three points! |
| Re: Nairaland Mathematics Clinic by Ojayyonline(m): 1:27am On Apr 13, 2015 |
Wow! I've been all out for dis kind of platform cos I hate seeing myself as d best around wen I'm sure ppl are better out thr. I'm just seeing dis today n I knw I'v missed a lot, I'm back online though just 4 dis reason. Keep it rolling guys!!! |
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