Author Topic: Maths & statistics exercices / French and English lessons  (Read 24845 times)

June 16, 2013, 03:58 PM
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Clearly, a section mathematics is missing on this forum. Some readers are probably not interested only in downloading, even though I fear too many are coming only to download and leave.
So I have a little problem: Here is a random variable X and its distribution probability:
P(X=0)=1/4, P(X=1)=1/12 P(X=2)=1/6 P(X=3)=1/2
Its Expected value is E(X)=23/12
Im looking for the standard deviation. I found 0,909 but a program gives me 1.26. I recall the standard deviation is given by sqrt(E(X²)-E(X)²) according to the formula of konig huygens.

It was an error, doing it again I found 1.26. thats ok.
« Last Edit: January 25, 2016, 11:59 AM by scarface »

August 01, 2013, 12:22 PM
Reply #1
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Here is an interesting biquadriatic equation,that some of you can eventually resolve:
x^4-(e²+1)x²+e²=0
tips: a change of variable is required. There are 2 couples of solutions.
« Last Edit: August 01, 2013, 04:16 PM by scarface »

August 01, 2013, 06:04 PM
Reply #2
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I'm glad you opened this topic. Many years ago when I was in school I was a whiz at math. Unfortunately it's been many years and I didn't get the opportunity to study higher math. I still remember quite a bit.

In this example you gave, naturally the idea is to factor the equation. Hmm... I'm familiar with regular quadratic equations but not biquadratics. Tell me more.

August 01, 2013, 07:24 PM
Reply #3
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Indeed,without e^2 at the end it coule be a factorisation. Its in fact a change of variable thats required: x^2=X. Then try to resolve X^2-(e^2+1)X+e^2=0. Ill give the solution later.

August 03, 2013, 05:03 AM
Reply #4
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here the solution:
So there is a classic polynomial to resolve: X²-(e²+1)X+e²=0
Delta=(e²+1)²-4e² and cant be used easily...But there is 1 obvious root: 1
We also know X2=c/X1 So We have the solutions:
X1=1, X2=e²/1=e²
Then we have to change the variable again to come back to the first equation:
as x²=X, x=sqrtX or x=-sqrtX
then x=sqrtX1=1 or x=sqrtX2=e or x=-sqrtX1=-1 or x=-sqrtX2=-e
So the equation x^4-(e²+1)x²+e²=0 has four solutions: {(1,e,-1,-e)}.

August 03, 2013, 02:14 PM
Reply #5
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Here is a harder equation to resolve: (cosx)^3-(sinx)^3=0

Here is the function f(x)=(cosx)^3-(sinx)^3
« Last Edit: November 21, 2020, 04:42 PM by scarface »

August 04, 2013, 12:08 PM
Reply #6
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Here is the solution:
We are going to study the function h(x)=(cosx)^3-(sinx)^3
h(x)=gof(x) with g(x)=x^3 and f(x)=cosx-sinx.
We know that x^3 is bijective and therefore injective and
by definition x^3=x'^3<=>x=x'.
Then (cosx)^3=(cosx')^3<=>cosx=cosx'
and  (sinx)^3=(sinx')^3<=>sinx=sinx'
then (cosx)^3-(sinx)^3=0 <=> cosx-sinx=0 <=> cosx=sinx
We know that cosx and sinx are 2pi periodic
and that cosx=sin pi/2 -x
then cosx=sinx <=> sin pi/2-x=sinx+2kpi with k a relative number
by identification pi/2-x=x+2kpi <=> 2x=pi/2+2kpi <=> x=pi/4+kpi
So there is an infinity of solutions with a pi periodicity.
« Last Edit: August 06, 2013, 03:19 AM by scarface »

August 06, 2013, 11:29 AM
Reply #7
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Probably it was a little hard. For the second exercice,i had the correction and for the third exercice i knew the method. Note that its essential to study the periodicity before the identification for any equation with cos, sin or tan. Clearly i dont deserve a field medal, the equivalent of nobels for mathematics. Im going to propose an exercise with vectors, the specialist humbert will appreciate.
If you have any problem of mathematics, feel free to ask.
« Last Edit: August 06, 2013, 11:56 AM by scarface »

August 07, 2013, 11:52 AM
Reply #8
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So I have renamed the topic maths exercices. Its better like that. Im now proposing an exercice with vectors. I dont have the correction, but its probably quite easy, a drunk college student could probably resolve it. the wording was in french so I translated it:

The space is linked to an orthonormal system (O; i; j). S is a Real number
We give the points A (8 ; 0 ; 8 ), B (10 ; 3 ; 10) and the line (D) defined by the parametric equations:
        x = − 5 + 3 s
       y = 1 + 2 s
       z = − 2 s         s= IR

1) a) give a system of parametric equations defined by the line Δ defined by A and B.   
b) Demonstrate that (D) and (Δ) are not coplanar. 
2) a) The plane (P) is parallel to (D) and contains (Δ). Demonstrate the vector n (2 ; − 2 ; 1) is normal to (P) then determine an cartesian equation of (P) .
b) Demonstrate that that the distance of any point M of (D) to (P) is independant of M.   
c) give a system of parametric equations of the line defined by the intersection of ( P) with the plane (xOy).
3) the sphere (S) is tangent to the plane ( P) on the point C(10 ; 1 ; 6). The center Ω of S is at the distance d = 6 of ( P) , on the same side that O. Give the cartesian equation of S . 

August 11, 2013, 01:42 PM
Reply #9
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So Im going to give a correction for the last exercice. Humbert has already revised its factorisation and is perhaps waiting impatiently to check his results. I did this one under notepad to copy it directly, but I couldn't do the last part which is difficult.

1)
a) The vector AB(2;3;2) is a direction vector of Δ.
Then,
 x = 8+2t 
 y = 3t    t=IR
 z = 8+2t
b)2 lines are coplanar if they are not parallel, nor secant.
let's take the vector u(3,2,-2) a direction vector of (D).
-lets prove they are not parallel:
AB(2,3,2) and u(3,2,-2) are not collinear (there is no real number k such as AB=ku)
then (Delta) and (D) are not parallel.
-lets prove they are not secant:
We have to resolve the system to eventually find a point of intersection:     
 x = - 5 + 3 s
 y = 1 + 2 s
 z = - 2 s
 x = 8+2t 
 y = 3t   
 z = 8+2t
 <=>
 -5+3s = -2s <=> s= 1
 y = 1 + 2 s
 z = - 2 s
 x = z 
 y = 3t   
 z = 8+2t
 <=>
 s = 1
 y = 3
 z = -2
 y = 3 <=> t = 1
 z = 10
 The system has no solution then the lines (AB) and Δ are not secant.
 The lines (AB) and Δ are not parallel nor secant, they are not coplanar.
2)
a) The plane (P) is parallel to (D) and contains Δ. The vector n (2 ; - 2 ; 1) is normal to (P).
Then the vectors AB(2,3,2) and u(3,2,-2) are direction vectors of the plane (P).
Let's do The scalar products to check if n is normal to (P):
n • u = 3 * 2 + 2 * (– 2) + (– 2) * 1 = 6 – 4 – 2 = 0
AB • n = 2 * 2 + 3 * (– 2) + 2 * 1 = 4 – 6 + 2 = 0
n is orthogonal to u and AB, and therefore orthogonal to (P), n is a normal vector to (P).
Knowing the normal vector, the cartesian equation of (P) has the form: 2x-2y+z+d=0
The plane (P) contains Delta and the point A(8,0,8) then
2*8-2*0+8+d=0 <=> d=-24
the cartesian equation of (P) is 2x-2y+z-24=0
b)lets take the point M(x,y,z).
we know the coordinates of M verify the system:
x = - 5 + 3 s
y = 1 + 2 s
z = - 2 s
Let's calculte the distance of M to (P) with the known formula:
d(M,(P))=|axM+byM+czM|/sqrt(a²+b²+c²)
then
d(M,(P))= |2x-2y+z-24|/sqrt(2²+(-2)²+1)=|2(-5+3s)-2(1+2s)-2s|/sqrt(9)=|-36|/3=12
We find a real number independent of the point M. Then the distance of any point M of (D)
to (P) is independent of M. The interpretation of this result is obvious since (D) is parallel to (P).
c) To give a parametric equation of (P) with (xOy) let's resolve the system:
2x-2y+z-24=0
z=0
<=>
2x=2y+24
z=0
<=>
x=y+12
z=0
The system of parametic equations is:
x=t+12
y=t      t=IR
z=0
3)

« Last Edit: August 11, 2013, 01:45 PM by scarface »