Unizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.
Unknown vector X(x1,x2) is defined as a linear combination of given non-collinear vectors A(a1,a2) and B(b1,b2): X = a·A + b·B.
Find the coefficients γ and δ in representation of the same vector X as a linear combination of another pair of given non-collinear vectors C(c1,c2) and D(d1,d2): X = γ·C + δ·D
Solution A
Straight forward method would be to express existing equations in coordinate form.
The following pair defines unknown x1 and x2 from equation X=a·A+b·B x1 = a·a1 + b·b1 x2 = a·a2 + b·b2
Then, using them as known variables, the second pair of equations defines unknown coefficients γ and β from equation X=γ·C+δ·D x1 = γ·c1 + δ·d1 x2 = γ·c2 + δ·d2
This is a simple algebraic problem of solving a system of two linear equations with two unknowns γ and δ.
Its solution is
γ =
d2·x1−d1·x2
c1·d2−c2·d1
δ =
c1·x2−c2·x1
c1·d2−c2·d1
where x1 and x2 were defined above in terms of a, b, a1, a2, b1, b2.
Problem B
Two vectors in two-dimensional space are presented on a picture below.
Find the cosine of an angle between them.
Solution B1
Recall that a scalar (dot) product of two vectors X and Y equals to a product of their magnitudes (moduli, lengths) by a cosine of an angle φ between them: X · Y = |X| · |Y| · cos(φ)
from which we can easily express cosine of an angle between these vectors in terms of their scalar product and magnitudes.
On the other hand, the same scalar product and magnitudes can be expressed in coordinate form, which can be easily derived from a picture.
Our two vectors in coordinate form are RED(4,−4) BLUE(2,4)
Their scalar product is RED · BLUE = 4·2 − 4·4 = −8
Their magnitudes are |RED| = 4√2 |BLUE| = 2√5
Now, dividing scalar product by a product of magnitudes, we can get a cosine of an angle between these vectors. cos(φ) = −8/(8√10) = −1/√10
Calculations show the approximate result arccos(−0.316227766) =
= 108.434949°
that is slightly more than 90°, which does look like on the picture above.
Solution B2
Let's solve this problem trigonometrically.
Start with bringing both vectors to the same beginning point - the origin of coordinates. The RED vector will have the coordinates of its tip X(4,−4), while the BLUE one will have the coordinates of its tip Y(2,4).
So, we have to determine a cos(∠XOY)
To simplify our work, consider point Z(4,4).
Obviously, ∠XOZ=90° and ∠XOY = ∠YOZ + 90°
In its turn, ∠YOZ = α − β
Knowing sin(α), cos(α), sin(β) and cos(β), we can easily evaluate any trigonometric function of α−β and, from it, any trigonometric function of ∠XOY=α−β+90°.
Length of OY is 2√5.
Therefore, sin(α) = 4/(2√5) = 2/√5 cos(α) = 2/(2√5) = 1/√5
Length of OZ is 4√2.
Therefore, sin(β) = 4/(4√2) = 1/√2 cos(β) = 4/(4√2) = 1/√2
From this sin(α−β) =
= sin(α)·cos(β) − cos(α)·sin(β) =
= (2/√5)·(1/√2) − (1/√5)·(1/√2) =
= 1/√10
Using trigonometric identity cos(φ+90°) = −sin(φ)
we derive cos(∠XOY) = cos(α−β+90°) =
= −sin(α−β) = −1/√10
which is exactly what we have calculated in Solution B1.
Given a two-dimensional vector A with known coordinates (a1,a2) and another two-dimensional vector B with known coordinates (b1,b2).
Find the coordinates (x,y) of vector X, if it's known that scalar productA·X = c and scalar productB·X = d.
Solution A
We have a system of two linear equations with two unknowns a1·x + a2·y = c b1·x + b2·y = d
Answer A x = (c·b2 − d·a2) / Δ y = (d·a1 − c·b1) / Δ
where Δ is determinant of the matrix of coefficients Δ = a1·b2 − a2·b1
Note A1
The obvious necessary and sufficient condition for the problem to have a unique solution is that Δ≠0, which means that vectors A and B must not be collinear.
Note A2
This problem can be easily extended to three-dimensional or any N-dimensional case with N-dimensional vectors A(a1,a2,...,aN) B(b1,b2,...,bN) X(x1,x2,...,xN)
and scalar products A·X = Σi∈[1,N]ai·xi B·X = Σi∈[1,N]bi·xi
The solution would lead to a system of three or, generally, N linear equations with as many unknowns - coordinates of unknown vector X.
Similar condition on the determinant of a matrix of coefficients not equal to zero should be specified to have a unique solution.
Problem B
Determine a three-dimensional vector X(x1,x2,x3)
if its known that X ⨯ A = B and A⊥X
where A(a1,a2,a3) and B(b1,b2,b3) are given non-collinear vectors and ⨯ signifies a vector (cross) product of two three-dimensional vectors.
Solution B
First of all, let's analyze the geometry of these three vectors A, B and X.
Since X ⨯ A = B, vector B, as follows from the definition of the vector product, must be perpendicular to two others: B⊥X B⊥A
Since it's given that A⊥X, all three vectors are mutually perpendicular to each other.
In particular, both A and X lie in the plane perpendicular to vector B.
The magnitude of vector B equals to a product of magnitudes of A and X times a sine of an angle from A to X.
The direction of vector B is along the perpendicular to a plane defined by vectors A and X.
Without the condition A⊥X we have the freedom to change the magnitude of X and rotate it within the same plane where it was, changing inverse-proportionally the sine of the angle from A to X and getting the same vector B=X⨯A. So, this last condition is important to get to the unique solution.
With the condition A⊥X the sine of an angle from A to X is 1 and all we need to do to obtain vector X is to determine its magnitude |X| from an obvious equation
|A|·|X| = |B|
since the direction of B is known - it's perpendicular to a plane defined by A and B.
Knowing the magnitude |X|=|B|/|A| and line perpendicular to a plane defined by A and B, along which vector X should be positioned, all that remains to establish is to chose one of two directions along this line. This is an elementary choice based on the right hand rule of forming a vector product.
Let's approach this problem in coordinate form.
Perpendicularity A⊥B can be expressed in terms of scalar product A·B=0, that is a1·b1 + a2·b2 + a3·b3 = 0
We will use it later on.
As derived in the Math 4 Teens course on UNIZOR.COM site (see lecture Vectors - Vector Product - Coordinate Form), since X⨯A=B, b1 = x2·a3 − x3·a2 b2 = x3·a1 − x1·a3 b3 = x1·a2 − x2·a1
This is a system of three linear equations with three unknowns.
Can we solve it and be done?
NO!
Our geometric analysis showed that without condition A⊥X the problem does not have a unique solution, and this condition is not part of our system of 3 linear equations with 3 unknowns.
What's wrong with solving the above system of 3 equations?
THESE THREE EQUATIONS ARE LINEARY DEPENDENT and, therefore, we really have only two independent equations, not three.
To determine this, we can rewrite our three equations in a canonical form x1·0 + x2·a3 + x3·(−a2) = b1 x1·(−a3) + x2·0 + x3·a1 = b2 x1·a2 + x2·(−a1) + x3·0 = b3
and calculate the determinant of a matrix of coefficients Ω:
Condition A⊥X will be the really independent third equation a1·x1+a2·x2+a3·x3 = 0
Let's assume that a3≠0, in which case we can find x1 and x2 in terms of x3 from the first two equations of the system above. b1 + x3·a2 = x2·a3 −b2 + x3·a1 = x1·a3
from which follows
[b1 + x3·a2]/a3 = x2
[−b2 + x3·a1]/a3 = x1
Substitute this into an expression a1·x1+a2·x2+a3·x3 = 0
getting a1·[−b2 + x3·a1]/a3 +
+ a2·[b1 + x3·a2]/a3 +
+ a3·x3 = 0
Resolve it for x3: x3·(a12+a22+a32) =
= a1·b2 − a2·b1
or
x3 = (a1·b2 − a2·b1)/ |A|2
where |A| is a modulus (magnitude, length) of vector A.
Obviously, we could've come to the same result if we considered the mutual perpendicularity of all three vectors A, B and X. This factor alone is sufficient to state that vector A⨯B is collinear with vector X.
As we know (see the reference above, when we considered B=X⨯A), vector Y=A⨯B has coordinates y1 = a2·b3 − a3·b2 y2 = a3·b1 − a1·b3 y3 = a1·b2 − a2·12
which exactly what numerators in the above expressions for X-components are.
The magnitude of vector Y is |Y|=|A|·|B|.
Since it's given that |B|=|A|·|X|, all we need to find vector X is to take vector Y and to alter its length to satisfy the condition |B|=|A|·|X|, which means to divide it by |A|.
Prove that function f(x) = x/(1+x²) is
(a) monotonously decreasing on interval (−∞,−1),
(b) monotonously increasing on interval (−1,1) and
(c) monotonously decreasing on interval (1,+∞).
Solution A
Our first observation is that, if we replace x with −x, function f(x)=x/(1+x²) changes the sign while retaining the absolute value f(−x) = −f(x).
That means, our function is odd, and its graph is centrally symmetrical relative to the origin of coordinates.
It also means that f(0)=0 and the graph on interval (−∞,0) can be obtained from the graph on interval (0,+∞) by turning it by 180° around the origin of coordinates.
When this rotation is done, an interval on the positive side of the X-axis, where the function was increasing, would be transformed by a rotation into a corresponding interval on the negative side, and the function there would still be increasing.
Similarly, an interval on the positive side of the X-axis, where the function was decreasing, would be transformed by a rotation into a corresponding interval on the negative side, where the function would still be decreasing.
The following is an illustration of this characteristic of odd functions.
Therefore, it's sufficient to investigate the behavior of this function on interval (0,+∞), and the results are sufficient to tell how this function behaves on interval (−∞,0).
If we prove that f(x) is monotonously increasing on interval (0,1) from f(0)=0 to f(1)=½, it implies that this function is monotonously increasing on interval (−1,0) from f(−1)=−½ to f(0)=0
Analogously, if we prove that f(x) is monotonously decreasing on interval (1,+∞) from f(1)=½ to infinitesimal value, it implies that this function is monotonously decreasing on interval (−∞,−1) from infinitesimal value to f(−1)=−½.
Let's prove that f(x) is increasing on interval (0,1).
Assume, both x1 and x2, which is greater than x1, belong to interval (0,1), and prove that f(x2) is greater than f(x1).
Analysis: x1/(1+x12) < x2/(1+x22)
Multiply both sides by positive value (1+x12)·(1+x12), which does not affect the inequality. x1·(1+x22) < x2·(1+x12)
Open the parenthesis, getting x1 + x1·x22 < x2 + x2·x12
Rearrange, x1·x2·(x2−x1) < x2 − x1
Since we assumed that both x1 and x2, which is greater than x1, belong to interval (0,1), all components of this inequality are positive, we can cancel the multiplier (x2−x1), arriving to an obvious inequality x1·x2 < 1
Proof:
All transformations of inequalities are invariant and reversible, therefore, from the obvious inequality x1·x2 < 1
we can obtain the original inequality using only invariant transformations, which proves its validity.
Let's prove that f(x) is decreasing on interval (1,+∞).
Assume, both x1 and x2, which is greater than x1, belong to interval (1,+∞), and prove that f(x1) is greater than f(x2).
Analysis: x1/(1+x12) > x2/(1+x22)
Multiply both sides by positive value (1+x12)·(1+x12), which does not affect the inequality. x1·(1+x22) > x2·(1+x12)
Open the parenthesis, getting x1 + x1·x22 > x2 + x2·x12
Rearrange, x1·x2·(x2−x1) > x2 − x1
Since we assumed that both x1 and x2, which is greater than x1, belong to interval (1,+∞), all components of this inequality are positive, we can cancel the multiplier (x2−x1), arriving to an obvious inequality x1·x2 > 1
Proof:
All transformations of inequalities are invariant and reversible, therefore, from the obvious inequality x1·x2 > 1
we can obtain the original inequality using only invariant transformations, which proves its validity.
Problem B
Prove that function f(x) = x + (1/x2) is
(a) monotonously increasing on interval (−∞,0),
(b) monotonously decreasing on interval (0,∛2) and
(c) monotonously increasing on interval (∛2,+∞).
Solution B1
Consider an expression
Δ = f(x2) − f(x1)
for x1 being smaller than x2 on each of the three intervals mentioned in the problem.
Based on the sign of Δ that reflects the relationship between f(x1) and f(x2), we can determine the behavior of the function on each interval.
(a) −∞ < x1 < x2 < 0
Δ = x2+(1/x22) − x1−(1/x12) =
= (x2−x1) + (1/x22−1/x12) =
= (x2−x1) +
+ (x1−x2)·(x1+x2)/(x12·x22) =
Since −∞ < x1 < x2 < 0, all components of the above expression are positive, which proves that in this interval the function is monotonously increasing.
(b) 0 < x1 < x2 < ∛2
Δ = (x2−x1) +
+ (x1−x2)·(x1+x2)/(x12·x22) =
= (x2−x1) ·
· [1 − (x1+x2)/(x12·x22)] =
= (x2−x1) ·
· (x1·x2−1/x1−1/x2)/(x1·x2)
The term (x2−x1) is positive.
The term (x1·x2) is positive.
Let's evaluate the sign of x1·x2−1/x1−1/x2 =
= x1·x2 − (1/x1+1/x2)
Component (x1·x2) on interval from 0 to ∛2 does not exceed ∛2·∛2 = ∛4.
Components (1/x1+1/x2) exceeds 1/∛2+1/∛2 = ∛4.
Therefore, in this case Δ is negative, which means that the function is decreasing in this interval.
(c) ∛2 < x1 < x2 < +∞
Evaluating again the term in question x1·x2 − (1/x1+1/x2)
Now situation with the sign of this term is opposite to the previous one.
Component (x1·x2) on this interval exceed ∛2·∛2 = ∛4.
Components (1/x1+1/x2) does not exceed 1/∛2+1/∛2 = ∛4.
Therefore, in this case Δ is positive, which means that the function is increasing in this interval.
Solution B2
Obviously, using the apparatus of Calculus, the solution is shorter.
The first derivative of the function f(x)=x+(1/x2) is f'(x) = 1 −2/x3
This derivative is equal to zero only at one point x = ∛2 = 21/3
The local minimum at that point is y = 21/3 + 2−2/3
When x is either negative or is greater then ∛2, the derivative is positive and, therefore, the function is increasing.
When x is positive but less than ∛2, the derivative is negative and, therefore, the function is decreasing.
Graph B
Incidentally, the graph of the function f(x)=x+(1/x2) looks like this
Solve the following equation X3+7·X2−21·X−27=0
if it's given that its three solutions form a geometric progression.
Solution A
Let's analyze the relationship between the solutions to an equation of the 3rd power X3+P·X2+Q·X+R=0
and its coefficients.
If a, b and c are the solutions of an equation of the 3rd power with a coefficient 1 at X3 then (X−a)·(X−b)·(X−c) = 0
Opening the parenthesis, we obtain X3−(a+b+c)·X2+
+(ab+bc+ca)·X−abc = 0
Notice that for this type of equation the product of its solutions equals to a free coefficient R with minus sign, while a sum of its solutions equals to a coefficient P at X2 with a minus sign.
Therefore, in case of a given equation a·b·c = −R = 27 and a + b + c = −P = −7
Since it's given that three solutions of this equation form a geometric progression, b = a·r and c = a·r²
where r is a progression's common ratio.
This allows us to represent two equations above in terms of a and r as follows a·(a·r)·(a·r²) = 27 a + (a·r) + (a·r²) = −7
From the first equation follows: (a·r)³ = 27 a·r = 3
which, incidentally, is one of the solutions: b=3.
From this and the above equations connecting solutions to coefficients of an equation, we conclude a·3·c = 27 and a + 3 + c = −7
or a·c = 9 and a + c = −10
Therefore expressing c in terms of a from the second equation and substituting into the first, we obtain c = −10 − a and a·(−10−a)=9
or a² + 10·a + 9 = 0
Solutions to this simple quadratic equations are −1 and −9.
Putting all solutions into a geometric sequence, a=−1, b=3, c=−9
or a=−9, b=3, c=−1
The common ratio of these progressions are r=−3 for the first and r=−1/3 for the second sequence, but the set of solutions is the same.
Answer A
Solutions to a given equation are: −1, 3, −9
They form a geometric progression with the first member a=−1 and the common ratio r=−3 or a geometric progression with the first member a=9 and the common ratio r=−1/3.
Problem B
Solve the following equation X3−15·X2+26·X+120=0
if it's given that its three solutions form an arithmetic progression.
Solution B
Let's use the same relationship between the solutions to an equation of the 3rd power as above a·b·c = −R = −120 and a + b + c = −P = 15
Since it's given that three solutions of this equation form an arithmetic progression, b = a+d and c = a+2d²
where d is a progression's common difference.
This allows us to represent two equations above in terms of a and d as follows a·(a+d)·(a+2d) = −120 a + (a+d) + (a+2d) = 15
From the second equation follows: 3a+3d = 15 a+d = 5
which, incidentally, is one of the solutions: b=5.
From this and the above equations connecting solutions to coefficients of an equation, we conclude a·5·c = −120 and a + 5 + c = 15
or a·c = −24 and a + c = 10
Therefore expressing c in terms of a from the second equation and substituting into the first, we obtain c = 10 − a and a·(10−a) = −24
or a² − 10·a − 24 = 0
Solutions to this simple quadratic equations are −2 and 12.
Putting all solutions into an arithmetic sequence, a=−2, b=5, c=12
The common difference of this progression is d=7.
Alternatively, using obvious relationships a = b − d and c = b + d,
we can express the equation a·5·c = −120
as (5−d)·5·(5+d) = −120
or 25 − d2 = −24
from which we derive d = ± 7,
which, in turn, leads to either a = −2, c = 12
or a = 12, c = −2
which is the same set of solutions in a different order.
Answer B
Solutions to a given equation are: −2, 5, 12
They form an arithmetic progression with the first member a=−2 and the common difference d=7 or an arithmetic progression with the first member a=122 and the common difference d=−7.
Problem C
Given two progressions, arithmetic and geometric, that satisfy the following conditions:
(a) the first members of these progressions are the same and are not equal to zero;
(b) the sums of the first three members of these progressions are the same;
(c) the sum of the first two members of arithmetic progression is greater than the sum of the first two members of geometric progression by the triple value of the first member of arithmetic progression.
What is the common ratio of the geometric progression?
Solution C
Arithmetic progression can be represented as
{a+k·d}
where a is the first member, d is the common difference, k∈[0,N] is a sequence number (N is greater than 3).
Geometric progression can be represented as
{b·rk}
where b is the first member, r is the common ratio, k∈[0,M] is a sequence number (M is greater than 3).
Conditions specified in the problem can be represented as
(a) a = b
(b) a + (a+d) + (a+2d) =
= b + b·r + b·r2
(c) a + (a+d) = 3a + b + b·r
Condition (a) can be used to replace unknown b in conditions (b) and (c) to simplify them as follows:
(b) a + (a+d) + (a+2d) =
= a + a·r + a·r2
(c) a + (a+d) = 3a + a + a·r
which can be simplified further as
(b) 2a + 3d = a·r·(1+r)
(c) d = 2a + a·r
Now we can use condition (c) to substitute d in condition (b):
(b) 2a + 3·(2a+a·r) = a·r·(1+r)
We can reduce this equation by a getting a quadratic equation for unknown ratio r: 2 + 3·(2+·r) = r·(1+r)
or r2 − 2r −8 = 0
with solutions r1=4; r2=−2
Answer C
There are two possible values for the common ratio r: r1=4; r2=−2
Note
We will use both terms average and mean interchangeably, but usually with proper qualification, like geometric mean or harmonic average.
Problem A
The Fibonacci sequence F(n) for integer n=0,1,2,3... is a sequence of numbers that starts with F(0)=0, F(1)=1
and satisfies the rule for any integer n≥0: F(n) + F(n+1) = F(n+2)
which results in F(2)=F(0)+F(1)=0+1=1 F(3)=F(1)+F(2)=1+1=2 F(4)=F(2)+F(3)=1+2=3 F(5)=F(3)+F(4)=2+3=5 F(6)=F(4)+F(5)=3+5=8 F(7)=F(5)+F(6)=5+8=13
etc.
Derive formula for the Fibonacci sequence.
Hint A
Prove that if the following equation Xn + Xn+1 = Xn+2
is true for n=0, it's true for any other positive integer n.
Solution A
Let's prove the statement in the Hint A above by using mathematical induction.
For n=0 it's assumed to be true.
Assume, it's true for some n, that is assume that Xn + Xn+1 = Xn+2
is true for some n.
We have to prove that it's true for n+1, that is we have to prove that Xn+1 + Xn+2 = Xn+3
Indeed, Xn+1 + Xn+2 =
= X·(Xn + Xn+1) =
= [use assumption for n] =
= X·Xn+2 = Xn+3
The formula Xn + Xn+1 = Xn+2
looks very much like the definition for Fibonacci sequence F(n) + F(n+1) = F(n+2)
But are there any X that satisfy the initial conditions for Fibonacci sequence F(0)=0, F(1)=1?
If yes, the solution for a formula describing the sequence would be solved.
The answer is YES.
First of all, let's find value(s) of X for n=0, that is let's solve the equation Xn + Xn+1 = Xn+2
for n=0. X0 + X1 = X2 or 1 + X = X2 or X2 − X − 1 = 0
It has two solutions X1,2 = (1 ± √1+4)/2 =
= (1 ± √5)/2
Both of these values, if used in the expression Xn + Xn+1 = Xn+2
transform it into identity for any natural n because they do it for n=0 and because we have proven above by induction that, if true for n=0, it's true for all natural n.
Let's check if a condition F(n) + F(n+1) = F(n+2)
is satisfied for F1(n)=X1n and F2(n)=X2n.
For X1: F1(n) = [(1+√5)/2]n F1(n+1) = [(1+√5)/2]n+1 F1(n) + F1(n+1) =
= [(1+√5)/2]n+[(1+√5)/2]n+1=
= [(1+√5)/2]n·[1+(1+√5)/2] =
= [(1+√5)/2]n·[(3+√5)/2] =
= [(1+√5)/2]n·[(6+2√5)/4] =
= [(1+√5)/2]n·[(1+2√5+5)/4] =
= [(1+√5)/2]n·[(1+√5)/2]2 =
= [(1+√5)/2]n+2 = F1(n+2)
Similarly, the condition F(n) + F(n+1) = F(n+2)
is satisfied for X2, we omit analogous calculations.
While the main condition of Fibonacci sequence, connecting any element with a sum of two preceding ones, is satisfied for both variants above, initial conditions for Fibonacci sequence F(0)=1, F(1)=1
are not yet satisfied.
Notice that if X1n + X1n+1 = X1n+2 and X2n + X2n+1 = X2n+2 then p·X1n+q·X2n +
+ p·X1n+1+q·X2n+1 =
= p·X1n+2+q·X2n+2
for any p and q.
So, the obvious suggestion is to find such p and q that a function F(n) = p·X1n+q·X2n
where X1=(1+√5)/2 and X2=(1−√5)/2, would satisfy the initial condition as well as the formula connecting the next sequence value with a sum of two preceding ones.
Condition F(0)=0 gives p·X10+q·X20 = 0
Any non-zero number raised to the power of 0 gives 1. So, this condition is equivalent to p + q = 0
Condition F(1)=1 gives p·X11+q·X21 = 1
Any non-zero number raised to the power of 1 is itself. So, this condition is p·X11 + q·X21 = 1 or p·(1+√5)/2 + q·(1−√5)/2 = 1
Now we have a system of two linear equations with two unknowns p and q.
From the first equation q = −p
Substitute it to the second equation: p·(1+√5)/2 − p·(1−√5)/2 = 1
from which follows p = 1/√5 and q = −1/√5
The final formula for Fibonacci sequence is, therefore, F(n) = (1/√5)·[(1+√5)/2]n −
− (1/√5)·[(1−√5)/2]n
Note
We will use both terms average and mean interchangeably, but usually with proper qualification, like geometric mean or harmonic average.
Problem A
Given that a fence around a rectangular field should have the length L, what should the sides of this field be to maximize the field's area?
What is the maximum area of a field in this case?
Hint
Arithmetic mean (average) of two positive real numbers is not less than their geometric mean (average).
These averages are equal only if the participating numbers are equal.
Answer
All sides must be equal to L/4 to form a square field.
The area will then be L²/16.
Problem B
In lecture UNIZOR.COM - "Math 4 Teens" - "Math Concepts" - "Induction" - "Averages" we have proven that arithmetic mean of N positive real numbers is greater or equal to their geometric mean: (X1+X2+...+XN)/N ≥
≥ (X1·X2·...·XN)1/N
Prove now that geometric mean of N positive numbers is greater or equal to their harmonic mean: (X1·X2·...·XN)1/N ≥
≥ N/(1/X1+1/X2+...+1/XN)
Hint B
Use the theorem about arithmetic and geometric means mentioned above for numbers 1/X1, 1/X2 etc.
Problem C
Prove that quadratic mean of N positive numbers is greater or equal to their arithmetic mean: sqrt[(X12+X22+...+XN2)/N] ≥
≥ (X1+X2+...+XN)/N
where sqrt is a square root function.
Hint C
The proof in general case is similar to the one with only a few numbers.
Let's analyze the situation with only 2 components.
Start from what's necessary to prove sqrt[(a2+b2)/2] ≥ (a+b)/2
Since all numbers are positive, get rid of sqrt function by raising both sides of an inequality to the power of 2 and apply invariant transformations. (a2+b2)/2 ≥ (a+b)2/4 2·(a2+b2) ≥ a2+2a·b+b2 a2+b2−2a·b ≥ 0 (a−b)2 ≥ 0
which is obvious, and all transformations are reversible.
Therefore, the proof is to derive the required inequality by reversed transformations from the obviously correct last inequality.
Let's illustrate the same approach for 4 numbers. sqrt[(a2+b2+c2+d2)/4] ≥
≥ (a+b+c+d)/4
Since all numbers are positive, get rid of sqrt function by raising both sides of an inequality to the power of 2. (a2+b2+c2+d2)/4 ≥
≥ (a+b+c+d)2/16
Use the invariant transformations: 4·(a2+b2+c2+d2) ≥
≥ a2+b2+c2+d2+
+2ab+2ac+2ad+
+2bc+2bd+2cd
3·(a2+b2+c2+d2) ≥
≥ 2ab+2ac+2ad+
+2bc+2bd+2cd
(a−b)2+(a−c)2+(a−d)2+
+(b−c)2+(b−d)2+(c−d)2 ≥ 0
which is obvious, and all transformations are reversible.
Therefore, the proof is to derive the required inequality by reversed transformations from the obviously correct last inequality.
Proof C
Given N positive real numbers Xi (i∈[1,N]).
Notice that
[Σ1≤ i ≤NXi]2 =
= Σ1≤ i ≤NXi2 +
+ Σ1≤ i < j ≤N2XiXj
Let's start from the obvious
Σ1≤ i < j ≤N(Xi−Xj)2 ≥ 0
From this follows: (N−1)·Σ1≤ i ≤NXi2 −
− Σ1≤ i < j ≤N2XiXj ≥ 0
Move products of different numbers to the right side of the inequality and add the sum of the squares of all numbers to both sides. N·Σ1≤ i ≤NXi2 ≥
≥ Σ1≤ i ≤NXi2 +
+ Σ1≤ i < j ≤N2XiXj
Transform the right side into a square of the sum of all numbers. N·Σ1≤ i ≤NXi2 ≥
≥ [Σ1≤ i ≤NXi]2
Summary
Harmonic mean (HM) N / [Σi∈[1,N]1/Xi]
is less or equal to geometric mean (GM)
[Πi∈[1,N]Xi]1/N
which is less or equal to arithmetic mean (AM)
[Σi∈[1,N]Xi] / N
which is less or equal to quadratic mean (QM) sqrt{[Σi∈[1,N]Xi2] / N}
Given two parallel lines and a segment AB on one of them.
Using only a straight line ruler, increase the length of segment AB by a factor of N.
In other words, find a point B' on the same line as AB such that the length of AB' is N times greater than the length of AB.
Solution A
Let's repeat the doubling of a segment, explained in the previous lecture Geometry 09 as Problem C.
Assume, AB is on the lower parallel line as on the picture below.
Choose any segment CD along the upper parallel line and divide it in halves by point P, as described in the Problem B of lecture Geometry 09.
Connect A and C, connect B and P. Lines AC and BP intersect at some pointM (if they don't and happened to be parallel, choose a longer CD.)
Now connect M and D and extend it to intersect with the lower parallel line at point B'.
The segments AB and BB' have equal length because segments CP and BP have equal length and two pairs of triangles are similar:
ΔAMB is similar to ΔCMP,
ΔBMB' is similar to ΔPMD.
Now we are ready to increase the length of AB by any factor.
To do this, just repeat the doubling of the size for segment BB' getting point B", so segment AB" has a triple length of AB.
Repeating this procedure any number of times we will get the new segment's length any number of times larger than the length of AB.
Problem B
Given two parallel lines and a segment AB on one of them.
Using only a straight line ruler, divide segment AB into N sub-segments of equal length.
In other words, find points B1, B2, ...,BN−1 on segment AB such that the length of any segment BiBi+1 is 1/Nth of the length of AB for any i∈[0,N−1], assuming A= B0 and B=BN.
Solution B
Assume, AB is on the lower parallel line.
Choose any segment CD along the upper parallel line and increase its length by a factor of N as described in the Problem A. Associate symbol D0 with point C, D1 with D and new points that double the length of a previous segment will be D2, D3, ...,DN.
Connect A=B0 with C=D0 and B with DN, extending these two lines to an intersection point M (if they are parallel and do not intersect, choose different point D.)
Connecting point M with each Di and extending to intersect with AB, gives all the points Bi.
