## Maths & statistics exercises / French and English lessons

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

#### scarface

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

#2
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)