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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:
agentofchange is that all about moment generating function

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:
@ factorial1 u try sir ,
differential calculus is quite cheep , but even though integral calculus is the reverse of it , its not usually funny to do so . Now in this regards , how can we then obtain back the integrand .?
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.

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Re: Nairaland Mathematics Clinic by agentofchange1(m): 2:18pm On Apr 12, 2015
jackpot:
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.

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:
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.
Cymbal:
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.
Cymbal:
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.
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:
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)
google functional equations
Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:47pm On Apr 12, 2015
jackpot:
google functional equations

try this dear let's see.
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:27pm On Apr 12, 2015
Cymbal:
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).

3:1

2 Likes

Re: Nairaland Mathematics Clinic by Cymbal: 9:42pm On Apr 12, 2015
agentofchange1:


3:1
kindly show the workings!
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:49pm On Apr 12, 2015
Cymbal:
kindly show the workings!

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:
Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.
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:
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.
thanks a bunch..
Re: Nairaland Mathematics Clinic by jackpot(f): 10:02pm On Apr 12, 2015
jackpot:
question 3

a military jet bomber flying at an altitude 600m in a horizontal straight line above a target with a speed of 360km/h drops off a bomb in order to hit the target with an initial velocity 0km/h. Suppose that at the time of drop, the horizontal distance of the target from the jet is 750m and the bomb's explosion radius is 300m, determine whether the target is hit by the bomb. If not, what is the range of the miss?
Re: Nairaland Mathematics Clinic by jackpot(f): 10:04pm On Apr 12, 2015
Laplacian:

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
another angle of attack, that's cool
Re: Nairaland Mathematics Clinic by Laplacian(m): 10:14pm On Apr 12, 2015
agentofchange1:


nice try sir, av known of this method too but the problem is , its still an approximate, since we truncate @ the 3rd term of expansion , why not 4th or 5th or higher ?. that's because we already know the answer to be 2 by trial -by-error , which is not always the ideal way of solving mathematical problems .


Now can it work for this ?

4^x = x^4 .?

or 3x = 8^x

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:
another angle of attack, that's cool

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:

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

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:


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
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:
Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.
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:
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.
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:

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!!
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
agentofchange1:
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)
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:
it helps buh please kindly give me full solution and explain how k=8/3.THANKS.
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:

heard of Newton's theorem for the radius of curvature of a line?
nah
Re: Nairaland Mathematics Clinic by Laplacian(m): 1:14am On Apr 13, 2015
Cymbal:
nah

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:


ok. Just google "intersecting arc theorem" because i can't draw the diagram here
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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