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Maths & statistics exercises / French and English lessons

Started by scarface, June 16, 2013, 11:58 PM

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scarface

Clearly, a maths section was 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.

First and foremost, do we have to say math or maths in English? Actually both are correct, however math is American English and math British English. Whether there is a s or not, the word is always singular.
Note that in French, the word "mathématiques" is always used in the plural, and you can also find the abbreviation "maths" and "math".


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.

scarface

#1
Here is an interesting biquadriatic equation,that some of you can eventually solve:
x^4-(e²+1)x²+e²=0
tips: a change of variable is required. There are 2 couples of solutions.

humbert

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.

scarface

#3
Quote from: humbert on August 02, 2013, 02:04 AMI'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.
Indeed,without e^2 at the end it could be a factorization. It's in fact a change of variable that is required: x^2=X. Then try to solve X^2-(e^2+1)X+e^2=0. Ill give the solution later.

scarface

#4
Here is the solution:
We try to solve x⁴-(e²+1)x²+e²=0, this is a biquadriatic equation.
Here, a change of variable is required:  x²=X.
We are now trying to solve X²-(e²+1)X+e²=0
So now we have a classic polynomial equation to solve: X²-(e²+1)X+e²=0

Here we have something that looks like this ax²+bx+c=0
Let's calculate the discriminant.
I remind you that the discriminant of a polynomial is a quantity that depends on the coefficients and determines various properties of the roots. It is generally defined as a polynomial function of the coefficients of the original polynomial. It is often denoted by the symbol Δ
The discriminant of the quadratic polynomial ax²+bx+c, with a ≠ 0 is: Δ = b² - 4ac
If Δ < 0  ax² + bx + c = 0 has no solution in R
If Δ = 0 then the equation has one solution x = −b/2a.
Si Δ >0 then the equation has 2 distinct solution x' and x": x' =( −b + √Δ )/2a and x" =( −b − √Δ )/2a.

Here, Δ=(e²+1)²-4e² and cant be used easily...
But there is 1 obvious root: 1 since 1²-(e²+1)*1+e²=0
What's more (e²+1)²-4e²=(e²)²+2e²+1-4e²=e⁴-2e²+1≃40.82. So Δ>0
And we know that for a polynomial ax²+bx+c tel with a≠0 Δ>0 and with x1 and x2 the 2 roots of the polynomial, then x2=c/(a*x1).
Here X1=1 (the obvious root).
So we know that X2=c/X1
Then 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)}.

scarface

#5
Here is a harder equation to solve: (cosx)^3-(sinx)^3=0

Here is the function f(x)=(cosx)^3-(sinx)^3

scarface

#6
Here is the solution:
(cosx)^3-(sinx)^3=0 <=> (cosx)^3=(sinx)^3
We are going to study the the 2 following functions:
h(x)=(cosx)^3
h'(x)=(sinx)^3

h(x)=gof(x) with g(x)=x^3 and f(x)=cos(x)
h'(x)=gof'(x) with g(x)=x^3 and f'(x)=sin(x)
We know that g is bijective and therefore injective, which means that if g(x)=g(x'), x=x'.
Then gof(x)=gof'(x)<=>f(x)=f'(x)
so (cosx)^3=(sinx)^3<=>(cosx)=(sinx)
We know that cos(x) and sin(x) are 2π periodic and that cos(x)=sin(π/2-x).
then cos(x)=sin(x) <=> sin(π/2-x)=sin(x+2kπ) with k a relative number.
by identification π/2-x=x+2kπ <=> 2x=π/2-2kπ <=> x=π/4+kπ (the sign before k is not important since it's a relative number).
So the solution is  x=π/4+kπ. There is an infinity of solutions with a π periodicity.

Note that if you have to do your research in French, the words in mathematics are very similar (bijection/bijective, fonction...)

scarface

#7
Probably it was a little hard. For the second exercise,i had the correction and for the third exercise 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 don't deserve a field medal, the equivalent of nobels for mathematics. I'm going to propose an exercise with vectors, humbert will appreciate.
If you have any problem of mathematics, feel free to ask.

scarface

#8
So I have renamed the topic maths exercises. It's better like that.
I'm now proposing an exercise with vectors. The wording was in french so I translated it:

The space is linked to an orthonormal system OIJ. 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+3s
y =1+2s
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 that the vector n (2 ; − 2 ; 1) is normal to (P) then determine a Cartesian equation of (P) .
b) Demonstrate that the distance of any point M of (D) to (P) is independent 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.


scarface

#9
So I'm going to give a correction for the last exercise. 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 solve 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 solve 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)