Proof by Mathematical Induction - Uplift Education...Proof by Mathematical Induction Principle of...
Transcript of Proof by Mathematical Induction - Uplift Education...Proof by Mathematical Induction Principle of...
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Proof by Mathematical Induction Principle of Mathematical Induction (takes three steps)
TASK: Prove that the statement Pn is true for all nβ π΅
1. Check that the statement Pn is true for n = 1. (Or, if the assertion is that the statement is true for n β₯ a, prove it for n = a.)
2. Assume that the statement is true for n = k (inductive hypothesis)
3. Prove that if the statement is true for n = k, then it must also be true for n = k + 1
Complex numbers
β π + ππ , πππ , πππ π = ββ1
β π§ = π + ππ π = π π(π§), π = πΌπ(π§)
β πππ£ππ π§ = π + ππ & π€ = π + ππ, π + ππ = π + ππ βΊ π = π & π = π
β πππ£ππ π§1 = πππ1 & π§2 = π
π π2 , π§1 = π§2 βΊ |π§1| = |π§2| & π1 = π2 + 2ππ. π = 0,Β±1,Β±2β¦
Presentation of complex number in Cartesian and polar coordinate system
π = π + ππβ πͺππππππππ ππππ
= π(πππ π½ + π πππ π½)β πππππ ππππ
πππ ππππβππππππππ ππππ
= ππππβπ¬ππππ ππππ
= π πππ π
Argan plane is the complex plane
βͺ ππππ = [|ππ|π
ππ½π] [|ππ|πππ½π] = |ππ||ππ|π
π(π½π+π½π)
βͺ ππ ππ= [|ππ|π
ππ½π]
[|ππ|πππ½π]=| ππ|
| ππ|ππ(π½πβπ½π)
βͺ ππππ’ππ’π ππ π΄ππ πππ’π‘π π£πππ’π: |π§| = π = βπ₯2 + π¦2
βͺ π΄πππ’ππππ‘: πππ π§ = π = πππ π‘ππ π¦
π₯ (ππππ ππππ‘π’ππ)
βͺ π₯ = ππππ π, π¦ = ππ ππ π
argument is measured in radians
Very useful for fast conversions from Cartesian into Euler form is β
ππ₯: π§ = β 2 π β π§ = 2ππ3π2
2 Properties of modulus and argument
βͺ |π§β| = |π§| & πππ (π§β) = βπππ π§
βͺ π§π§β = |π§|2
βͺ πππ = ππ(π+π2π) πππ
De Moivreβs Theorem
π§π = [π(πππ π + π π πππ)]π = ππ(πππ π + π π πππ)π
π§π = [ππππ]π= ππππππ = ππ(πππ ππ + π π ππ ππ)
β΄ (πππ ππ½ + π πππ ππ½) = (ππππ½ + π ππππ½)π
Finding n-th root of a complex number z π§ = |π|πππ = |π|ππ(π+π2π)
βπ§π
= β|π§|π
ππ (π+π2π)
π = β|π§|π
{πππ (π + π2π
π) + π π ππ (
π + π2π
π) } π = 0, 1, 2, 3β¦ , π β 1
Only the values k = 0, 1, β¦, n - 1 give different values of βπ§π
. There are exactly n nth roots of z.
Geometrically, the n th roots are the vertices
of a regular polygon with n sides in Argan plane.
β1
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Reminder and Conjugate roots of polynomial equations with real
coefficients Real Polynomials A real polynomial is a polynomial with only real coefficients Two polynomials are equal if and only if they have the same degree (order) and corresponding terms have equal coefficients. if 2x3 + 3x2 β 4x + 6 = ax3 + bx2 β cx + d then
a = 2, b = 3, c = β 4, d = 6
Remainder
If P(x) is divided by D(x) until a remainder R is obtain:
π(π₯)
π·(π₯)= π(π₯) +
π (π₯)
π·(π₯)
π(π₯) = π·(π₯)π(π₯) + π (π₯)
famous nth root of unity is the solution of equation : π§π = 1
D(x) is the divisor Q(x) is the quotient R(x) is the remainder
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When we divide by a polynomial of degree 1 ("ax+b") the remainder will have degree 0 (a constant)
From here to :
The Remainder Theorem
When a polynomial P(x) is divided by (x β k) until a remainder R is obtain, then R = P(x):
proof:
π(π₯)
π₯ β π= π(π₯) +
π
π₯ β π
π(π₯) = π(π₯)(π₯ β π) + π
π(π) = π
The Factor Theorem
k is a zero of P(x) β (x β k) is a factor of P(x) proof: k is a zero of P(x) β P(k) = 0
β R = 0
as π(π₯) = π(π₯)(π₯ β π) + π = π(π₯)(π₯ β π) =
β (x β k) is a factor of P(x)
Corollar
(x β k) is a factor of P(x) β there exists a polynomial Q(x) such that P(x) = (x β k) Q(x)
Properties of Real Polynomials
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Polynomials: Sums and Products of Roots
Let p and q be roots of quadratic equation ax2 + bx + c=0
π + π = βπ/π ππ = π/π
Let p, q and r be roots of quadratic equation ax3 + bx2 + cx + d = 0
π + π + π = βπ/π πππ = βπ/π
Vectors, Lines and Planes
Vector as position vector of point A in three dimensions in Cartesian coordinate system:
οΏ½βοΏ½ = ππ₯π Μ+ ππ¦πΜ + ππ§οΏ½ΜοΏ½ β‘ (
ππ₯ππ¦ππ§) β‘ (ππ₯ ππ¦ ππ§)
Both, position vector of point A and point A have the same coordinates:
οΏ½βοΏ½ = (
ππ₯ππ¦ππ§) , π΄ = (ππ₯ , ππ¦ , ππ§)
π,Μ πΜ πππ οΏ½ΜοΏ½ are unit vectors in x, y and z directions.
π Μ = (100) πΜ = (
010) οΏ½ΜοΏ½ = (
001)
|οΏ½βοΏ½| = βππ₯2 + ππ¦
2 + ππ§2 |οΏ½βοΏ½| ππ ππππππ ππππππ‘π’ππ, πππππ‘β, ππππ’ππ’π ππ ππππ
Unit vector A unit vector is a vector whose length is 1. It gives direction only!
οΏ½ΜοΏ½ = οΏ½βοΏ½
|οΏ½βοΏ½|= ππ₯ οΏ½ΜοΏ½ + ππ¦πΜ + ππ§οΏ½ΜοΏ½
βππ₯2 + ππ¦
2 + ππ§2
=1
βππ₯2 + ππ¦
2 + ππ§2
(
ππ₯ππ¦ππ§)
Vector between two points
π΄π΅ββββ ββ = (
π₯π΅ β π₯π΄π¦π΅ β π¦π΄π§π΅ β π§π΄
) = (π₯π΅ β π₯π΄) πΜ + (π¦π΅ β π¦π΄) πΜ + (π§π΅ β π§π΄) οΏ½ΜοΏ½
π΅π΄ββββ ββ = (
π₯π΄ β π₯π΅π¦π΄ β π¦π΅π§π΄ β π§π΅
) = (π₯π΄ β π₯π΅) π Μ + (π¦π΄ β π¦π΅) πΜ + (π§π΄ β π§π΅) οΏ½ΜοΏ½
ππππ’ππ’π β‘ πππππ‘β: |π΄π΅ββββ ββ | = |π΅π΄ββββ ββ | = β(π₯π΅ β π₯π΄)2 + (π¦π΅ β π¦π΄)
2 + (π§π΅ β π§π΄)2
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Parallel and Collinear Vectors
οΏ½βοΏ½ ππ ππππππππ π‘π οΏ½ββοΏ½ β οΏ½βοΏ½ = ποΏ½ββοΏ½ πππ πππππ‘π πππ πππππππππ ππ π‘βππ¦ πππ ππ π‘βπ π πππ ππππ
π΄, π΅ πππ πΆ πππ πππππππππ β π΄π΅ββββ ββ = ππ΄πΆββββββ πππ π πππ π πππππ π
(πππ ππππππ πππππ‘ πππ π‘βπ π πππ ππππππ‘πππ)
The Division of a Line Segment
X divides [AB]β‘ ABΜ Μ Μ Μ in the ratio π: π means π΄πββββββ : ππ΅ββ ββ ββ = π : π
Internal Division: External Division: P divides [AB] internally in ratio 1:3. Find P X divide [AB] externally in ratio 2:1. Find Q
π΄πββββββ : ππ΅ββββββ = 1: 3 β π΄πββββββ =1
4 π΄π΅ββββ ββ π΅πββ ββ ββ = π΄π΅ββββ ββ
Dot / Scalar Product (Scalar, can be Β± )
οΏ½βοΏ½ β’ οΏ½ββοΏ½ = οΏ½ββοΏ½ β’ οΏ½βοΏ½ = |οΏ½βοΏ½||οΏ½ββοΏ½| cos π
Properties of dot product
οΏ½βοΏ½ β’ οΏ½ββοΏ½ = οΏ½ββοΏ½ β’ οΏ½βοΏ½
ππ οΏ½βοΏ½ πππ οΏ½ββοΏ½ πππ ππππππππ, π‘βππ οΏ½βοΏ½ β’ οΏ½ββοΏ½ = |οΏ½βοΏ½||οΏ½ββοΏ½|
ππ οΏ½βοΏ½ πππ οΏ½ββοΏ½ πππ πππ‘πππππππππ, π‘βππ οΏ½βοΏ½ β’ οΏ½ββοΏ½ = β |οΏ½βοΏ½||οΏ½ββοΏ½|
οΏ½βοΏ½ β’ οΏ½βοΏ½ = |οΏ½βοΏ½ |2
(οΏ½βοΏ½ + οΏ½ββοΏ½) β’ (π + π) = οΏ½βοΏ½ β’ π + οΏ½βοΏ½ β’ π + οΏ½ββοΏ½ β’ π + οΏ½ββοΏ½ β’ π
οΏ½βοΏ½ β’ οΏ½ββοΏ½ = 0 (οΏ½βοΏ½ β 0, οΏ½ββοΏ½ β 0) β οΏ½βοΏ½ πππ οΏ½ββοΏ½ πππ ππππππππππ’πππ
πΜ β’ π Μ = 1 πΜ β’ πΜ = 1 οΏ½ΜοΏ½ β’ οΏ½ΜοΏ½ = 1 & πΜ β’ πΜ = 0 πΜ β’ οΏ½ΜοΏ½ = 0 πΜ β’ οΏ½ΜοΏ½ = 0
In Cartesian coordinates: οΏ½βοΏ½ β’ οΏ½ββοΏ½ = (
ππ₯ππ¦ππ§)(
ππ₯ππ¦ππ§
) = ππ₯ππ₯ + ππ¦ππ¦ + ππ§ππ§
Cross / Vector Product (vector) The magnitude of the vector οΏ½βοΏ½ Γ οΏ½ββοΏ½ is equal
to the area determined by both vectors.
|οΏ½βοΏ½ Γ οΏ½ββοΏ½| = |οΏ½βοΏ½||οΏ½ββοΏ½| π ππ π
Direction of the vector οΏ½βοΏ½ Γ οΏ½ββοΏ½ is given by right hand rule.
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Properties of vector/cross product
οΏ½βοΏ½ Γ οΏ½ββοΏ½ = β οΏ½ββοΏ½ Γ οΏ½βοΏ½
ππ οΏ½βοΏ½ πππ οΏ½ββοΏ½ πππ ππππππππππ’πππ, π‘βππ |οΏ½βοΏ½ Γ οΏ½ββοΏ½| = |οΏ½βοΏ½||οΏ½ββοΏ½|
(οΏ½βοΏ½ + οΏ½ββοΏ½) Γ (π + π) = οΏ½βοΏ½ Γ π + οΏ½βοΏ½ Γ π + οΏ½ββοΏ½ Γ π + οΏ½ββοΏ½ Γ π
οΏ½βοΏ½ Γ οΏ½ββοΏ½ = 0 (οΏ½βοΏ½ β 0, οΏ½ββοΏ½ β 0) β οΏ½βοΏ½ πππ οΏ½ββοΏ½ πππ ππππππππ
πΉππ ππππππππ π£πππ‘πππ π‘βπ π£πππ‘ππ πππππ’ππ‘ ππ 0.
i Γ i = j Γ j = k Γ k = 0 i Γ j = k j Γ k = i k Γ i = j
β οΏ½βοΏ½ Γ οΏ½ββοΏ½ = (
ππ¦ππ§ β ππ§ππ¦ππ§ππ₯ β ππ₯ππ§ππ₯ππ¦ β ππ¦ππ₯
) = |
πΜ πΜ οΏ½ΜοΏ½ππ₯ ππ¦ ππ§ππ₯ ππ¦ ππ§
|
How do we use dot and cross product
β’ To find angle between vectors the easiest way is to use dot product, not vector product.
β The angle between two vectors
π = ππππππ οΏ½βββοΏ½ β’ οΏ½βββοΏ½
|οΏ½βββοΏ½||οΏ½βββοΏ½| Angle between vectors can be acute or obtuse
β The angle between two lines
π = ππππππ |οΏ½βββοΏ½ β’ οΏ½βββοΏ½|
|οΏ½βββοΏ½||οΏ½βββοΏ½| Angle between lines is by definition acute
οΏ½βοΏ½ πππ οΏ½ββοΏ½ πππ ππππππ‘πππ π£πππ‘πππ
βͺ Dot product of perpendicular vectors is zero.
βͺ To show that two lines are perpendicular use the dot product with line direction vectors.
β The angle between a line and a plane
π ππ π = πππ π =|οΏ½ββοΏ½ β’ π|
|οΏ½ββοΏ½||π|
π = πππ sin|οΏ½ββοΏ½ β’ π|
|οΏ½ββοΏ½||π|
β The angle between two planes is the same as the angle between their two normal vectors
π = πππ πππ |οΏ½ββοΏ½ β’ οΏ½βββοΏ½|
|οΏ½ββοΏ½||οΏ½βββοΏ½|
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βͺ To show that two planes are perpendicular use the dot product on their normal vectors.
βͺ To find all vectors perpendicular to both οΏ½βοΏ½ πππ οΏ½ββοΏ½ ππππ k (οΏ½βοΏ½ Γ οΏ½ββοΏ½), π π π
βͺ To find the unit vector perpendicular to both οΏ½βοΏ½ πππ οΏ½ββοΏ½ ππππ οΏ½ββοΏ½Γ οΏ½ββοΏ½
|οΏ½ββοΏ½Γ οΏ½ββοΏ½|
Coplanar four points
πΉππ’π πππππ‘π πππ ππππππππ ππ πππ ππππ¦ ππ π‘βπ π£πππ’ππ ππ π‘βπ π‘ππ‘ππβπππππ πππππππ ππ¦ π‘βππ ππ 0:
Volume of a tetrahedron = 1
6 scalar triple product
π =1
6|π β ( οΏ½βοΏ½ Γ οΏ½ββοΏ½)| =
1
6 β
ππ₯ ππ¦ ππ§ππ₯ ππ¦ ππ§ππ₯ ππ¦ ππ§
β π’πππ‘π 3
Lines A plane is completely determined by two points, what can be translated into a fixed point A and one direction vectors
β Vector equation of a line
The position vector οΏ½ββοΏ½ of any general point P on the line passing through point A
and having direction vector οΏ½ββοΏ½ is given by the equation
π = (π1πΜ + π2πΜ + π3οΏ½ΜοΏ½) + π(π1οΏ½ΜοΏ½ + π2πΜ + π3οΏ½ΜοΏ½) ππ (π₯π¦π§) = (
π1π2π3) + π(
π1π2π3
)
β Parametric equation of a line β Ξ» is called a parameter Ξ» β π
(π₯π¦π§) = (
π1π2π3) + π (
π1π2π3
) β
π₯ = π1 + ππ1π¦ = π2 + ππ2π§ = π3 + ππ3
β Cartesian equation of a line
π₯ = π1 + ππ1 βΉ π = (π₯ β π1)/π1
π¦ = π2 + ππ2 βΉ π = (π¦ β π2)/π2
π§ = π3 + ππ3 βΉ π = (π§ β π3)/π3
βΉ π₯βπ1
π1 =
π¦βπ2
π2 =
π§βπ3
π3 (= π)
β Distance from a point P to a line L
Point Q is on the line hence its coordinates must satisfy line equation.
ππππ£π πππ’ππ‘πππ ππββ ββ ββ β’ οΏ½ββοΏ½ = 0 πππ π. πΉπππ π‘βπππ ππππ πππππππππ‘ππ ππ πππππ‘ π πππ π π’ππ πππ’πππ‘ππ¦ |ππ|
β Relationship between 3 β D lines
β the lines are coplanar (they lie in the same plane). They could be:
βͺ intersecting βͺ parallel βͺ coincident
β the lines are not coplanar and are therefore skew (neither parallel nor intersecting)
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Are the lines β the same?β¦β¦.check by inspection
β parallel?β¦β¦β¦check by inspection
β skew or do they have one point in common?
solving π1βββ β = π2ββββ will give 3 equations in and Β΅.
Solve two of the equations for and Β΅.
if the values of and Β΅ do not satisfy the third equation then the lines are skew, and they do not intersect.
If these values do satisfy the three equations then substitute the value
of or Β΅ into the appropriate line and find the point of intersection.
β Distance between two skew lines π = οΏ½βοΏ½ + π οΏ½ββοΏ½ πππ π = π + π π
π = |οΏ½ΜοΏ½ β’ (π β οΏ½βοΏ½)| π€βπππ οΏ½ΜοΏ½ = οΏ½ββοΏ½ΓοΏ½βοΏ½
|οΏ½ββοΏ½ΓοΏ½βοΏ½|
Planes
A plane is completely determined by two intersecting lines, what can be translated into a fixed point A and two nonparallel direction vectors
β Vector equation of a plane
The position vector οΏ½ββοΏ½ of any general point P on the plane passing through point A
and having two direction vector οΏ½ββοΏ½ is given by the equation
π = οΏ½βοΏ½ + ποΏ½ββοΏ½ + ππ ππ (π₯π¦π§) = (
π1π2π3) + π(
π1π2π3
)
β Parametric equation of a plane β Ξ», π is called a parameters Ξ», π β π
(π₯π¦π§) = (
π1π2π3) + π(
π1π2π3
) + π (
π1π2π3) β
π₯ = π1 + ππ1 + ππ1π¦ = π2 + ππ2 + ππ2π§ = π3 + ππ3 + ππ3
β Normal/Scalar product form of vector equation of a plane
π β’ οΏ½ββοΏ½ = οΏ½βοΏ½ β’ οΏ½ββοΏ½ ππ οΏ½ββοΏ½ β’ (π β οΏ½βοΏ½) = 0
β Cartesian equation of a plane
π1π₯ + π2π¦ + π3π§ = π π = π1π1 + π2π2 + π3π3
To convert a vector equation into a Cartesian equation, you find the cross product of the two vectors appearing in the vector equation to find a normal to the plane and use that to find the Cartesian equation. To convert Cartesian -> vector form, you need either two vectors or three points that lie on the plane!
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π₯ + 3π¦ β 2π§ = 5 3π₯ + 5π¦ + 6π§ = 7 2π₯ + 4π¦ + 3π§ = 8
So first step is to choose three arbitrary random non-collinear points.
β Distance from origin
π· = |π β’ οΏ½ΜοΏ½| = |οΏ½βοΏ½ β’ οΏ½ΜοΏ½| = |π1π1+π2π2+π3π3 |
βπ12+π2
2+π32
β Intersection of a line and a plane
πΏπππ πΏ: π = (1β2β1) + π (
456) πππ πππππ π±: π₯ + 2π¦ + 3π§ = 5
To check if the line is parallel to the plane do dot product of direction vector of the line and normal vector to the plane. If dot product is not zero the line and the plane are not parallel and the line will intersect the plane in one point.
Substitute the line equation into the plane equation to obtain the value of the line parameter Β΅.
(1 + 4π) + 2(β2 + 5π) + 3(β1 + 6π) = 5 1 + 4Β΅ β 4 + 10Β΅ β 3 + 18Β΅ = 5 β π =11
32
Substitute Β΅ into the equation of the line to obtain the co-ordinates of the point of intersection:
1
32(76β934)
In general: Solve for Β΅ and substitute into the equation of the line to get the point of intersection. If this equation gives you something like 0 = 5, then the line will be parallel and not in the plane, and if the equation gives you something like 5 = 5 then the line is contained in the plane.
System of Linear Equations
Consider the following linear system :
THE A LINEAR SYSTEM MAY BEHAVE IN ANY OF THREE POSSIBLE WAYS 1. The system has a single unique solution. 2. The system has no solution. 3. The system has infinitely many solutions. For three variables, each linear equation is a plane in 3-D space, and the solution set is the intersection of these planes. Thus the solution set may be a plane, a line, a single point, or the empty set.
~ [1 3 β23 5 62 4 3
|578]
The augmented matrix
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Example: solution is a single point: Combine two at the time to eliminate one variable
1. step:
(1. πππ’ππ‘πππ) β (β3) + (2. πππ’ππ‘πππ) = πππ€ π πππππ πππ’ππ‘πππ (ππ π₯ πππ¦ ππππ)
(1. πππ’ππ‘πππ) β (β2) + (3. πππ’ππ‘πππ) = πππ€ π‘βπππ πππ’ππ‘πππ (ππ π₯ πππ¦ ππππ)
2. step
(2. πππ’ππ‘πππ) β (β1
2) + (3. πππ’ππ‘πππ) = πππ€ π‘βπππ πππ’ππ‘πππ (ππ π¦ πππ¦ ππππ)
β2π₯ β 6π¦ + 4π§ = β10β3π₯ β 9π¦ + 6π§ = β15
2π¦ β 6π§ = 4
π₯ + 3π¦ β 2π§ = 5 3π₯ + 5π¦ + 6π§ = 7 2π₯ + 4π¦ + 3π§ = 8
~ π₯ + 3π¦ β 2π§ = 5 β4π¦ + 12π§ = β8 β2π¦ + 7π§ = β2
~ π₯ + 3π¦ β 2π§ = 5 β4π¦ + 12π§ = β8 π§ = 2
This is equivalent to say that augmented matrix is in reduced row echelon form β augmented matrix with the zeroes in the bottom left corner.
[1 3 β23 5 62 4 3
|578] ~ [
1 3 β20 β4 120 β2 7
|5β8β2] ~ [
1 3 β20 β4 120 0 1
|5β82] ~
π₯ + 3π¦ β 2π§ = 5 β4π¦ + 12π§ = β8 π§ = 2
solution::
π = π π² = π π = βππ
Generally, using this method we get augmented matrix of coefficients of the system in echelon form
Echelon form
[π π π0 π π0 0 β
|πππ] ~
ππ₯ + ππ¦ + ππ§ = π ππ¦ + ππ§ = π βπ§ = π
β π§ =π
β β π¦ & π₯
ππππππ ππππππππ: β β 0 β π§ =
π
β β π¦ & π₯ (π πππ¦ ππ πππ¦ πππ‘ ππ 0)
ππ ππππππππ: β = 0 πππ π β 0 β 0 β π§ = π β 0 π€βππβ ππ πππ π’ππ β
π‘βπππ ππ ππ π πππ’π‘πππ (π π¦π π‘ππ ππ ππππππππππππ)
ππππππππππ ππππ πππππππππ: β = 0 πππ π = 0 β π§ πππ ππ πππ¦ ππ’ππππ, π π π€π π€πππ‘π:
π§ = π‘ π€βπππ π‘ β π
π₯ = π₯(π‘)
π¦ = π¦(π‘)
π§ = π‘
Parametric representation of infinitely many solutions is not unique. We expressed the variable which was not free (z) in terms of the parameter. We eliminated from row 3 variables x and y. It does not have to be that way. We could eliminate z and y to get x, and then express y and z. One can solve for any of the variables. Of course, the solution set will look different. However, it will still represent the same solutions.
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Example:
π’ππππ’π π πππ’π‘πππ: [1 1 10 2 20 0 3
|126] β π§ = 2 π¦ = β1 π₯ = 0 π(0, β1,2)
ππ π πππ’π‘πππ: [1 1 10 2 20 0 0
|123] β π§ β 0 = 3 πππ π’ππ
πππππππ‘πππ¦ ππππ¦ π πππ’π‘ππππ : [1 1 10 2 20 0 0
|120] β π§ β 0 = 0 β
π‘ππ’π πππ πππ¦ π§
β π§ = π‘ π π π¦ = 1 β π‘ π₯ = 0
Counting principles, including permutations and combinations.
The binomial theorem: expansion of (π + π)π, π πΊ π΅.
If there are π different ways of performing an operation and for each of these there are π different ways of performing a second independent operation, then there are ππ different ways of performing the two operations in succession. The product principle can be extended to three or more successive operations. The number of different ways of performing an operation is equal to the sum of the different mutually exclusive possibilities.
The word πππ suggests multiplying the possibilities
The word ππ suggests adding the possibilities.
If the order doesn't matter, it is a Combination.
If the order does matter it is a Permutation.
A permutation of a group of symbols is any arrangement of those symbols in a definite order.
Explanation: Assume you have n different symbols and therefore n places to fill in your arrangement. For the first place, there are n different possibilities. For the second place, no matter what was put in the first place, there are n β 1 possible symbols to place, for the rth place there are n β r +1 possible places until the point where r = n, at which point we have saturated all the places. According to the product principle, therefore, we have n (n β 1)(n β 2)(n β 3)β―1 different arrangements, or n!
β THE PRODUCT RULE
β COUNTING PATHS
β PERMUTATIONS (order matters)
β Permutations of π different object : π!
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Wise Advice: If a group of items have to be kept together, treat the items as one object. Remember that there may be permutations of the items within this group too.
Suppose we have 10 letters and want to make groups of 4 letters. For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. Good logic to apply to similar questions straightforward!!!
There are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multiplying each time.)
It is the number of ways of choosing π objects out of π available given that βͺ The order of the elements does not matter. βͺ The elements are not repeated [such as lottery numbers (2,14,15,27,30,33)]
The easiest way to explain it is to: assume that the order does matter (ie permutations), then alter it so the order does not matter.
Since the combination does not take into account the order, we have to divide the permutation of the total number of symbols available by the number of redundant possibilities. π selected objects have a number of redundancies equal to the permutation of the objects π! (since order doesnβt matter) However, we also need to divide the permutation n! by the permutation of the objects that are not selected, that is to say (π β π)! .
π!
π! (π β π)!
(ππ) β‘ πͺπ
π β‘ πͺππ =
π!
π! (π β π)!=π(π β π)(π β π)(π β π + π)
π!
β Binomial Expansion/Theorem
(π + π)π = β(ππ) ππβπππ = ππ + (
π1) ππβ1π +β―+ (
ππ) ππβπππ +β―+ ππ
π
π=0
β Permutations of π different objects out of π different available : π β (π β π) βββ (π β π + π)
β Permutations with repetition of π different objects out of π different available = ππ
β COMBINATIONS (order doesnβt matters)
π π π (π0) β‘ 1 0! β‘ 1
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β Binomial Coefficient
(ππ) is the coefficient of the term containing ππβπππ in the expansion of (π + π)π
(ππ) =
π(π β 1)(π β 2)β― (π β π + 1)
π!=
π!
π! (π β π)!=
π!
(π β π)! π!= (
ππ β π
)
πβπ πππππππ π‘πππ ππ (π + 1)π‘β π‘πππ ππ : ππ+1 = (ππ) ππβπππ
The constant term is the term containing no variables. When finding the coefficient of π₯π always consider the set of all terms containing π₯π
Probability
The number of trials is the total number of times the βexperimentβ is repeated.
The outcomes are the different results possible for one trial of the experiment.
Equally likely outcomes are expected to have equal frequencies.
The sample space, U, is the set of all possible outcomes of an experiment.
And event is the occurrence of one particular outcome.
π(π΄) =π(π΄)
π(π)
Complementary Events
Two events are described as complementary if they are the only two possible outcomes. Two complementary events are mutually exclusive.
Since an event must either occur or not occur, the probability of the event either occurring or not occurring must be 1.
π·(π¨) + π·(π¨β²) = π
Use when you need probability that an event will not happen
Possibility when we are interested in more than one outcome (events are βandβ, βorβ, βat leastβ)
Combined Events
βͺ (π’ππππ) β‘ πππ‘βππ β© (πππ‘πππ πππ‘πππ) β‘ πππ‘β/πππ
Given two events, B and A, the probability of at least one of the two events occurring,
π(π΄ βͺ π΅) = π(π΄) + π(π΅) β π(π΄ β© π΅)
P(A) is the probability of an event A occurring in one trial,
n(A) is the number of times event A occurs in the sample space
n(U) is the total number of possible outcomes.
either A or B or both
P(A) includes part of B from intersection P(B) includes part of A from intersection
π(π΄ β© π΅) (both A and B) was counted twice, so one has to be subtracted
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πΌπ‘ ππ ππππππ‘πππ‘ π‘π ππππ€ βππ€ π‘π πππ‘ π(π΄ β© π΅)
For mutually exclusive events (no possibility that A and B occurring at the same time)
Turning left and turning right (you can't do both at the same time) Tossing a coin: Heads and Tails
π(π΄ βͺ π΅) = π(π΄) + π(π΅) π(π΄ β© π΅) = β For non - mutually exclusive we are going to find conditional probability for
Independent and Dependent Events
A bag contains three different kinds of marbles: red, blue and green. You pick the marble twice. Probability of picking up the red one (or any) the second time depends weather you put back the first marble or not. β’ Independent Events: β’ Dependent Events:
the probability that one event occurs probability of one event occurring influences in no way affects the probability of the likelihood of the other event the other event occurring.
You put the first marble back You donβt put the first marble
β Conditional Probability:
Given two events, B and A, the conditional probability of an event A is the probability that the event will occur given the knowledge that an event B has already occurred. This probability is written as (notation for the probability of A given B)
P (A|B )
Probability of the intersection of A and B (both events occur) is: π(π΄ β© π΅) = π(π΅)π(π΄|π΅)
β’ Independent Events: β’ Dependent Events: π(π΄|π΅) = π(π΄) = π(π΄|π΅β²) π(π΄ β© π΅) = π(π΅)π(π΄|π΅)
π΄ ππππ πππ‘ ππππππ ππ π΅ πππ ππ π΅β² π(π΄|π΅) πππππ’πππ‘ππ πππππππππ ππ π‘βπ ππ£πππ‘ π΅
π(π΄ β© π΅) = π(π΄)π(π΅) π(π΄ β© π΅) = π(π΅)π(π΄|π΅)
π(π΄|π΅) =π(π΄ β© π΅)
π(π΅)
β Use of Venn diagrams, tree diagrams and tables of outcomes to solve problems.
1. Venn Diagrams
The probability is found using the principle π(π΄) =π(π΄)
π(π)
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2. Tree diagrams
A more flexible method for finding probabilities is known as a tree diagram.
This allows one to calculate the probabilities of the occurrence of events, even where trials are non-identical (where π(π΄|π΄) β π(π΄)), through the product principle.
β§ͺ Bayesβ Theorem
π(π΄ β© π΅) = π(π΅)π(π΄|π΅) βΉ π(π΄ β© π΅) = π(π΄)π(π΅|π΄)
π(π΄|π΅) = π(π΄ β© π΅)
π(π΅)=π(π΄)π(π΅|π΄)
π(π΅) π΅ππ¦ππ β² π‘βπππππ
βͺ Another form of Bayesβ theorem (Formula booklet)
From tree diagram: there are two ways to get A, either after B has happen or after B has not happened:
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π(π΄) = π(π΅)π(π΄|π΅) + π(π΅β²)π(π΄|π΅β²) βΉ π(π΅|π΄) =π(π΅)π(π΄|π΅)
π(π΅)π(π΄|π΅) + π(π΅β²)π(π΄|π΅β²)
βͺ Extension of Bayesβ Theorem
If there are more options than simply B occurs or B doesnβt occur, for example if there were three possible outcomes for the first event B1, B2, and B3
Probability of A occurring is: π(π΅1)π(π΄|π΅1) + π(π΅2)π(π΄|π΅2) + π(π΅3)π(π΄|π΅3)
π(π΅π|π΄) =π(π΅π)π(π΄|π΅π)
π(π΅1)π(π΄|π΅1) + π(π΅2)π(π΄|π΅2) + π(π΅3)π(π΄|π΅3)
Outcomes B1, B2, and B3 must cover all the possible outcomes.
Descriptive Statistics
Concepts of population, sample, random sample and frequency distribution of discrete and continuous data.
A population is the set of all individuals with a given value for a variable associated with them.
A sample is a small group of individuals randomly selected (in the case of a random sample) from the population as a whole, used as a representation of the population as a whole.
The frequency distribution of data is the number of individuals within a sample or population for each value of the associated variable in discrete data, or for each range of values for the associated variable in continuous data.
Presentation of data: frequency tables and diagrams
Grouped data: mid-interval values, interval width, upper and lower interval boundaries, frequency histograms.
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Mid interval values are found by halving the difference between the upper and lower interval boundaries.
The interval width is simply the distance between the upper and lower interval boundaries.
Frequency histograms are drawn with interval width proportional to bar width and frequency as the height.
Median, mode; quartiles, percentiles.
Range; interquartile range; variance, standard deviation.
Mode (discrete data) is the most frequently occurring value in the data set.
Modal class (continuous data) is the most frequently occurring class.
Median is the middle value of an ordered data set.
For an odd number of data, the median is middle data.
For an even number of data, the median is average of two middle data.
Percentile is the score bellow which a certain percentage of the data lies.
Lower quartile (Q1) is the 25th percentile.
Median (Q2) is the 50th percentile.
Upper quartile (Q3) is the 75th percentile.
Range is the difference between the highest and lowest value in the data set.
The interquartile range is Q3βQ1.
Cumulative frequency is the frequency of all values less than a given value.
The population mean, ΞΌ is generally unknown but the sample mean, οΏ½Μ οΏ½ used to serve as an unbiased estimate of this mean. That used to be. From now on for the examination purposes, data will be treated as the population. Estimation of mean and variance of population from a sample is no longer required.
A variable X whose value depends on the outcome of a random process is called a random variable. For any random variable there is a probability distribution/ function associated with it.
β Probability distribution/ function
Discrete Random Variables
P(X = x), the probability distribution of x, involves listing P(π₯π ) for each π₯π. 1. 0 β€ π(π = π₯) β€ 1
2. βπ(π = π₯) = 1
π₯
3. π(π = π₯π)
= 1 ββ π(π = π₯π)
πβ π
[π (ππ£πππ‘ π₯π ππππ’ππ ) = 1 β π(πππ¦ ππ‘βππ ππ£πππ‘ ππππ’ππ )]
Discrete and Continuous Random Variables
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Discrete Random Variables X defined on a β€ x β€ b
probability density function (p.d.f.), f (x), describes the relative likelihood for this variable to take on a given value cumulative distribution function ( c.d.f.), πΉ(π‘), is found by integrating the p.d.f. between the minimum value of X and t
πΉ(π‘) = π(π β€ π‘) = β«π(π₯)ππ₯
π‘
π
1. π(π₯) β₯ 0 πππ πππ π₯ π (π, π)
2. β« π(π₯) = 1
π
π
3. πππ πππ¦ π β€ π < π β€ π, π(π < π < π) = β« π(π₯)ππ₯
π
π
βͺ πΉππ π ππππ‘πππ’ππ’π ππππππ π£πππππππ, π‘βπ ππππππππππ‘π¦ ππ πππ¦ π πππππ π£πππ’π ππ π§πππ
π(π = π) = 0 β π(π β€ π β€ π) = π(π < π < π) = π(π β€ π < π) ππ‘π.
Expected value (or mean) is a weighted average of the possible values that X can take, each value
being weighted according to the probability of that event occurring.
Discrete Random Variables Continuous Random Variables
π¬(πΏ) = π = βπ π·(πΏ = π) π¬(πΏ) = π = β« π π(π)π π
β
ββ
Properties of expected value π¬(πΏ)
1. πΌπ π πππ π πππ ππππ π‘πππ‘π , π‘βππ πΈ(ππ + π) = ππΈ(π) + π 2. πΈ(π + π ) = πΈ(π) + πΈ(π )
Expected Value of a Function of X
πΈ[ π(π)], π€βπππ π(π) ππ π ππ’πππ‘πππ ππ π ππ :
Discrete Random Variables Continuous Random Variables
πΈ[ π(π)] = βπ(π₯)π(π = π₯) πΈ[ π(π)] = β« π(π₯)π(π₯)ππ₯
sum of: [(each of the possible outcomes)
Γ (the probability of the outcome occurring)].
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Mode is the most likely value of X.
Discrete Random Variables
The mode is the value of x with largest π(π = π₯) which can be different from the expected value
Continuous Random Variables The mode is the value of x where f(x) is maximum (which may not be unique).
Median
Discrete Random Variables
The median of a discrete random variable is the "middle" value. It is the value of X such that
π(π β€ π₯) β₯1
2 πππ π(π β₯ π₯) β₯
1
2
Continuous Random Variables
The median of a random variable X is a number m such that
β« π(π₯) =1
2
π
π
The median m is the number for which the probability is exactly Β½ that the random variable will have a value greater than m, and Β½ that it will have a value less than m.
Variance
The variance of a random variable tells us something about the spread of the possible values of the
variable. Variance, Var( X ), is defined as the average of the squared differences of X from the mean:
π½ππ(πΏ) = ππ = π¬ (πΏ β π)π = π¬(πΏπ) β ππ
Discrete Random Variables
π½ππ(πΏ) = π΄π₯2 π(π = π₯) β π2
Continuous Random Variables
π½ππ(πΏ) = β«π₯2π(π₯)ππ₯ β π2
π
π
Properties of variance
Note that the variance does not behave in the same way as expectation when we multiply and add constants to random variables. In fact:
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π ππ(ππ + π) = π2πππ(π)
πΌπ πππππππ πππ[π + π] β πππ(π) + πππ(π) (only for independent β not IB)
Standard deviation of X
π = βπππ(π)
Binomial Distribution - Discrete There are three criteria that must be met in order for a random probability distribution to be a binomial
distribution.
The experiment consists of n repeated trials.
Each trial can result in just two possible outcomes. We call one of these outcomes a success and
the other, a failure.
The probability of success, denoted by p, is the same on every trial.
The trials are independent; that is, the outcome on one trial does not affect the outcome on other
trials β the probability of success is a constant in each trial
If a random variable X has a binomial distribution, we write
β’ πΏ ~ π©(π, π) (~ means βhas distributionβ¦β).
and the probability density function is:
β’ π(π = π₯) = (ππ₯)ππ₯(1 β π)πβπ₯ π₯ = 0, 1, β¦ , π
β’ n is number of trials
β’ p is the probability of a success
β’ (1 β p) is the probability of a failure.
If X is a random variable is binomial with parameters n and p, then mean and variance are: β’
πΈ(π) = π = ππ
β’ πππ(π) = π2 = ππ(1 β π)
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Poisson Distribution
A discrete random variable X with a probability distribution function (p.d.f.) of the form:
β’ π(π = π₯) =ππ₯πβπ
π₯!, π₯ = 0, 1, 2, β¦
is said to be a Poisson random variable with parameter m. We write β’ π ~ ππ(π)
Mean and Variance
πΌπ π ~ ππ(π), π‘βππ β’ πΈ(π) = π
β’ πππ(π) = π Random Events The Poisson distribution is useful because many random events follow it. If a random event has a mean number of occurrences m in a given time period, then the number of occurrences within that time period will follow a Poisson distribution. For example, the occurrence of earthquakes could be considered to be a random event. If there are 5 major earthquakes each year, then the number of earthquakes in any given year will have a Poisson distribution with parameter 5.
There are three criteria that must be met in order for a random probability distribution to be a
Poisson distribution.
The average number of occurrences (m) is constant for every interval.
The probability of more than one occurrence in a given interval is very small.
The number of occurrences in disjoint intervals are independent of each other.
Sums of Poissons
Suppose X and Y are independent Poisson random variables with parameters β and m respectively. Then X + Y has a Poisson distribution with parameter β + m. In other words: If πΏ ~ π·π(π΅), and π ~ π·π(π), ), then πΏ + π~ π·π(π΅ +π)
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Normal distribution. Standardization of normal variables.
A continuous random variable X follows a normal distribution if it has the following probability density function :
β’ π(π₯) =1
πβ2π πβ12(π₯βππ)2
β β < π₯ < β
The parameters of the distribution are ΞΌ and ππ, where ΞΌ is the mean (expectation) of the distribution and π2 is the variance.
πΏ ~ π΅(π, ππ) the random variable X has a normal distribution with parameters ΞΌ and π2.
Properties
1. ππ
ππ₯ π(π₯) = β
1
π2β2π(π₯ β π
π) π
β12(π₯βππ)2
= 0 β πππ₯πππ’π : π = π
2. ππ
ππ₯ π(π₯) = β
1
π3β2π[1 β (
π₯ β π
π)2
] πβ12(π₯βππ)2
= 0 β πππππππ‘ππππ : π = π Β± π
The curve is symmetrical about the line x = ΞΌ
βͺ limπ₯βΒ±β
π(π₯) = 0
βͺ β« π(π₯)
β
ββ
= 1
βͺ π = πππ₯{π(π₯)}
For a normal curve, standard deviation Ο is uniquely determined as the horizontal distance from the
vertical line of symmetry π₯ = π to the point of inflection.
In a normal distribution, 68.26% of values lie within one standard deviation of the mean, 95.4% of values lie within two standard deviations of the mean and 99.74% of values lie within three standard deviations of the mean.
Standard score or z β score is the number of standard deviations from the mean.
The Standard Normal Distribution
If π ~ π(0, 1), then Z is said to follow a standard normal distribution. P(Z < z) is known as the cumulative distribution function of the random variable Z. For the standard normal distribution, this is usually denoted by F(z). Normally, you would work out the c.d.f. by doing some integration. However, it is impossible to do this for the normal distribution and so results have to be looked up in statistical tables.
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Standardising: Now, the mean and variance of the normal distribution can be any value and so clearly there can't be a statistical table for each one. Instead, we convert to the standard normal distribution- we can also use statistical tables for the standard normal distribution to find the c.d.f. of any normal distribution. Cumulative Distribution Functions is defined as π(π β€ π₯) and that what we need The standard normal distribution, or Z-distribution, is the application of the transformation
π =π β π
π
to a normal X-distribution, such that the mean is at x = 0 and there is one standard deviation per unit on the x-axis. Where the probability density function for normal distribution has two parameters and , the Z-distribution has none. This makes it useful when comparing results from two or more different normal distributions, since comparing Z-values allows one to take into account the standard deviation and mean when comparing results. Finding probabilities with a GDC involves using normalcdf(a,b,ΞΌ ,Ο ) for lower limit a and upper limit b (under βDISTRβ). It is important to note that (π β€ π) = π(π < π) .
24
CALCULATOR
Binomial Distribution β’ πΏ ~ π©(π, π)
β’ π(π = π₯) = (ππ₯)ππ₯(1 β π)πβπ₯ π₯
= 0, 1, β¦ , π
β’ n is number of trials
β’ p is the probability of a success
β’ (1 β p) is the probability of a failure.
β’ πΈ(π) = π = ππ β’ πππ(π) = π2 = ππ(1 β π)
π(π = π₯ ) = ππππππππ (π, π, π₯) π(π β€ π₯) = ππππππππ(π, π, π₯)
Poisson Distribution β’ π ~ ππ(π)
β’ π(π = π₯) =ππ₯πβπ
π₯!, π₯ = 0, 1, 2, β¦
β’ πΈ(π) = π
β’ πππ(π) = π π(π = π₯ ) = ππππ π πππππ (π, π₯) π(π β€ π₯) = ππππ π πππππ(π, π₯)
Normal distribution. β’ π ~ π(π, π2)
β’ π(π₯) =1
πβ2π πβ12(π₯βππ)2
β β < π₯ < β
π(π β₯ π₯ ) = πππππππππ ( π₯, 1πΈ99, π, π) π(π₯1 β€ π β€ π₯2) = πππππππππ ( π₯1, π₯2, π, π)
Standardized normal distribution. β’ π ~ π(0,1)
π(π β€ π§ ) = πππππππππ (β1πΈ99, π₯) π(π§1 β€ π β€ π§2) = πππππππππ ( π§1, π§2) When given (π β€ π) = π₯ πππ£ππππ(π₯, π, π) β’ π ~ π(π, π2)
πππ£ππππ(π§) β’ π ~ π(0,1) π(π β€ π) = π(π < π) πππππ’π π π(π = π) = 0
NormalCDF(lower , upper , mean , SD) o Gives probability that a value is within a given range
BinomCDF(trials , probability of event , value) o Gives cumulative probability, i.e. the number of successes within n trials is at most the value
BinomPDF(trials , probability of event , value) o Gives the probability for a particular number of success in n trials
PoissonCDF(mean , value) o Gives cumulative probability, i.e. probability of at most (value) occurrences within a time period
PoissonPDF(mean , value) o Gives probability of a particular number of occurrences within a time period
invNorm(percentage) o Given a probability, gives the corresponding z-score, i.e. standard deviations from the mean