Wednesday, March 28, 2007

Precalculus 10.5 Parametric Equations

10.5 Parametric Equations

check out: http://colalg.math.csusb.edu/~devel/precalcdemo/param/src/param.html

Up to this point, you have been representing a graph by a single equation involving two variables such as x and y. In this section, you will study situations in which it is useful to introduce a third variable to represent a curve in the plane.

Let's consider the path followed by an object that is propelled into the air at an angle of 45 degrees. If the initial velocity of the object is 48 feet per second, it can be shown that the object follows the parabolic path

y = -x2/72 + x

However, this equation does not tell the whole story. Although it does tell us where the object has been, it doesn't tell us when the object was at a given point (x,y) on the path. To determine this time, you can introduce a third variable t, which is called a parameter. It is possible to write both x and y as functions of t to obtain the parametric equations.

Example: x = 24Ö(2) t and y = -16t2 + 24Ö(2) t

Curvilinear motion: two variables for position, one for time.

A curve in the plane is said to be parameterized if the set of coordinates on the curve, (x,y), are represented as functions of a variable t. Namely, x = f(t), y = g(t).

Conversely, given a pair of parametric equations with parameter t, the set of points (f(t), g(t)) form a curve in the plane.

For this particular motion problem, x and y are continuous functions of t, and the resulting path is a plane curve. (Recall that a continuous function is one whose graph can be traced without lifting the pencil from the paper).

Definition of a Plane Curve:

If f and g are continuous functions of t on an interval I, the set of ordered pairs (f(t), g(t)) is a plane curve C. The equations

x= f(t) and y = g(t)

are parametric equations for C, and t is the parameter.

One way to sketch a curve represented by a pair of parametric equations is to plot points in the xy-plane. Each set of coordinates (x,y) is determined from a value chosen for the parameter t. By plotting the resulting points in the order of increasing values of t, you trace the curve in a specific direction. This is called the orientation of the curve.

The orientation of a parameterized curve is the direction determined by increasing values of the parameter. Sometimes arrows are drawn on the curve to denote the orientation.

Example 1: you are given a pair of parametric equations. Select 6 valid values of t and substitute each into both equations to get the coordinates of a point in the plane. Plot all six points in the plane.

x = t2 - 4 and y = t/2, where -2 ≤ t ≤ 3

t = -2, x = 0, y = -1
t = -1, x = -3, y = -.5
t = 0, x = -4, y = 0
t = 1, x = -3, y = .5
t = 2, x = 0, y = 1
t = 3, x = 5, y = 1.5

By plotting these points in the order of increasing t, you obtain the curve that does not define y as a function of x. The graph looks like a parabola that opens to the right with t = 0, x = -4, and y = 0 as its vertex.

to graph this in your TI-83 plus calculator, go to mode, arrow down to the line with func and arrow over to par (for parametric). Then press enter. Your calculator is now in parametric mode. Go to the y=. For X1T = T2 -4, Y1T = T/2. Now go to window, Tmin = 0, Tmax = 10, Tstep = 1, Xmin = -10, Xmax = 10, Xscl = 1, Ymin = -10, Ymax = 10, and Yscl = 1. Press 2nd graph and you will see the table above. Now press graph and you can see this curve.

Sometimes you can have two different sets of parametric equations that have the same graph. For example, x = 4t2 - 4 and y = t, -1 ≤ t ≤ 1.5

When you plot these, they have the same graph as the one above but the values of t are different. This means that the second graph is traced out more rapidly than the first. Therefore, in applying this to real situations, different parametric representations can be used to represent various speeds at which objects travel along a given path.

Eliminating the Parameter:

A second technique to identifying the curve of the parametric equations is to try to eliminate the parameter from the equations. This will result in an equation involving only x and y which we may recognize. For example, let's look again at an example.

Given x = 3 - 2t and y = 2 + 3t

Sketching this graph, we have the following data:

t = -2, x = 7, y = -4
t = -1, x = 5, y = -1
t = 0, x = 3, y = 2
t = 1, x = 1, y = 5
t = 2, x = -1, y = 8

then solving for t in the first equation

x = 3 - 2t
x - 3 = -2t
t = (x-3)/(-2)

Substitute for t into the second equation to get

y = 2 + 3t
y = 2 + 3((x-3)/(-2)
y = 2 + (3x - 9)/(-2)
y = (3x)/(-2) + 13/2

Now sketch this graph

x = 7, y = -4
x = 5, y = -1
x = 3, y = 2
x = 1, y = 5
x = -1, y = 8

We can sketch its graph using function graphing techniques. Be careful, however, to take into account any restrictions on the value of the parameter.

Notice anything about the two separate parts?

Converting equations from parametric to rectangular form can change the ranges of x and y. In such cases, you should restrict x and y in the rectangular equation so that its graph matches the graph of the parametric equations.

Example: Given x =Ö t and y = 1 - t
1st solve for t in the first equation:

x =Ö t
x2 = t with the constraint that x ³ 0

Therefore, y = 1 - t
y = 1 - x2

Parametric Equation:
t = 0, x = 0, y = 1
t = 1, x = 1, y = 0
t = 2, x = Ö2, y = -1
t = 3, x = Ö3, y = -2
t = 4, x = 2, y = -3

Now the rectangular Equation:
x = 0, y = 1
x = 1, y = 0
x = Ö2, y = -1
x = Ö3, y = -2
x = 2, y = -3

As you can see when you graph these, you have to restrict x ³

Example: Sketch the curve represented by

x = 4 + 2 cosq and y = 1 + 2 sinq

Begin by solving for cosq and sinq

x = 4 + 2 cosq
x - 4 = 2 cosq
(x - 4)/2 = cosq

y = 1 + 2 sinq
y - 1 = 2 sinq
(y - 1)/2 = sinq

Now make use of the identity sin2q + cos2q = 1

((y-1)/2)2 + ((x-4)/2)2 = 1

(y-1)2/4 + (x-4)2/4 = 1

(x-4)2 + (y-1)2 = 4

This is the graph of a circle with center point (4,1) and radius of 2

Now you can graph this putting your calculator in parametric mode

X1 = 4 + 2 cosT
Y1 = 1 + 2 sinT

Now your window Tmin = 0
Tmax = either 2p radians or 360 degrees depending whether you are in degrees or radians.
Tstep = play with this - try 90 degrees, 30 degrees, 1 unit and see what happens to your graph

Xmin = 0, Xmax = 7, Xscl = 1
Ymin = -3, Ymax = 4, Yscl = 1

Example: Find 2 different sets of parametric equations for the given rectangular equation:

y = x2

a). Let x = t so then using substitution y = t2

t = 0, x = 0, y = 0
t = 1, x = 1, y = 1
t = 2, x = 2, y = 4
t = 3, x = 3, y = 9
t = 4, x = 4, y = 16

b). Let x = t2 then using substitution y = t4

t = 0, x = 0, y = 0
t = 1, x = 1, y = 1
t =2, x = 4, y = 16

The x-values and the y-values are the same, but the t-values are faster.

You can think of this as a track race. In theory, the runners are following the same path. But they all go at a different speed. Therefore their location placements are the same (the x-values and y-values) but the time is different. Some slower and some quicker.

Saturday, March 24, 2007

Precalculus 6.4 Vectors and Dot Products

6.4 Vectors and Dot Products

The Dot Product of Two Vectors

The symbol of Dot Product is "• " and written like: uv
and spoken like the Dot Product of vector u and vector v

Definition of a Dot Product:
The Dot Product of u = < u1, u2> and v = <v1, v2 > is:
uv = (u1)( v1) + ( u2)(v2)

1. u • v = v • u
2. 0 • v = 0
3. u • (v + w) = uv + uw
4. vv = ll v ll 2
5. c (uv) = cuv = u • cv

Example 1: Find the dot product
< 2, 3 >< -1, 4 > = (2)(-1) + (3)(4) = -2 + 12 = 10

Example 2: u = < 5, 12 > and v = < -3, 2 >
uv = (5)(-3) + (12)(2) = -15 + 24 = 9

Example 3: u = < 2, 2 > and v = < -3, 4 > and w = < 1, -4 >
a) ll u ll - 2 = √(22+2 2) - 2 = √(4 + 4 ) - 2 = √(8) - 2 = 2√2 - 2

b) w • (u + v) = w • u + w • v
= ((1)(2) + (-4)(2)) + ((1)(-3) + (-4)(4))
= (2 + -8) + (-3 + -16 ) = - 6 + -19 = -25

c) (wu ) v = ((1)(2) + (-4)(2)) v
= (2 + -8) v
= (-6) v
= (-6) < -3, 4 >
= < 18, -24 >

d) 4uv = 4 (uv)
= 4 ((2)(-3) + (2)(4))
= 4 ( -6 + 8)
= 4 (2) = 8

Use the Dot Product to find ll u ll
u = < 2, -4 >
ll u ll 2 = u · u = (2)(2) + (-4)(-4) = 4 + 16
ll u ll = √20

ll u ll = √(( 2 2 +(-4)2 )= √(4 + 16) = √20

As you can see, it gives you the same answer!

The angle between two non-zero vectors is the angle θ, 0π ≤ θ ≤ π
between its respective standard position vectors. This angle can be found using
the dot product.

If θ is the angle between two nonzero vectors u and v, then

(ll u ll )(llvll )(Cos θ) = u • v

Check out this web site:
http://mathproofs.blogspot.com/2006/07/dot-product-and-cosine.html

Example : Find the angle θ between the vectors u and v where u = < 4, 4 > and v = < -2, 0 >
(ll u ll )(ll v ll )(Cos θ) = u • v
(ll< 4, 4 > ll )(ll< -2, 0 >ll )(Cos θ) = < 4, 4 >< -2, 0 >
(√(42 + 42)) √((-2)2 + (0)2) (Cos θ) = (4)(-2) + (4)(0)
(√(16 + 16)) √(4 ) (Cos θ) = -8
√(32) √(4 ) (Cos θ) = -8
(4√2)(2) (Cos θ) = -8
Cos θ = -√(2)/2
θ = -45 degrees or -π/4 and since the angle has to be in the first or second quadrant and cosine is negative, this would be the second quadrant or
θ = 3π/4

Definition of Orthogonal Vectors

The vectors u and v are orthogonal if u • v = 0
The terms orthogonal and perpendicular mean essentially the same thing - meeting at right angles.

Example : Are the vectors u = < 3, -3 > and v = < -1, -1 > orthogonal, parallel, or neither?
u • v = (3)(-1) + (-3)(-1) = -3 + 3 = 0 so they are orthogonal

Example: u = j and v = i - 2j so u = < 0, 1 > and v = < 1, -2 >
u • v = (0)(1) + (1)(-2) = 0 + -2 = -2 so they are not orthogonal

m1 = 1/0 and m2 = -2/1 so they are not parallel so they are neither

Example 3: u = < 3, 3 > and v = < 5, 5 >
u • v = (3)(5) + (3)(5) = 15 + 15 = 30 so they are not orthogonal

m1 = 3/3 = 1 and m2 = 5/5 = 1 so they are parallel

Precalculus 6.3b Direction Angles

6.3b Direction Angles

If u is a unit vector such that θ is the angle (measured counterclockwise) from the positive x-axis to u, the terminal point of u lies on the unit circle and you have

u = < x,y > = < cos θ, sin θ > = (cos θ) i + (sin θ) j

Suppose that u is a unit vector with direction angle θ. If v is any vector that makes an angle θ with the positive x-axis, then it has the same direction as u and you can write:

v = llvll < cos θ, sin θ > = ll v ll (cos θ) i + ll v ll(sin θ) j

Because v = ai + bj = ll v ll(cos θ) i + ll v ll(sin θ) j

it follows that the direction angle θ for v is determined from

tan θ = (sin θ) ÷ (cos θ)

= (ll v llsin θ) ÷ (ll v ll cos θ)

= b÷ a

Example 1: Find the direction Angle of the vector u = 2i + 4j

The direction angle is tan θ = b ÷ a = 4 ÷ 2 = 2

so tan-1(2) = θ
therefore θ = 63.43494882° and since this vector is in the first quadrant it is the angle.

Example 2: Find the magnitude and direction angle of the vector v = 8 (cos 135°i + sin 135°j )

v = 8 cos 135°i + 8 sin 135°j
recall v = ai + bj = ll v ll(cos θ) i + llv ll(sin θ ) j

therefore ll v ll = 8 and θ = 135°

Example 3: find the component form of the sum of u and v with direction angle θ u and θv

ll u ll = 2 and θu = 30 ° and ll v ll = 2 and θ v = 90 °

Vector u = 2 cos 30 ° i + 2 sin 30 ° j

=
2 (√(3)/2) i + 2(1/2) j = √(3) i + j

u = < √3 , 1 >

Vector v = 2 cos 90 ° i + 2 sin 90 ° j

=
2(0) i + 2(1) j = 2 j

v = < 0 , 2 >

u + v = < √3 , 1 > + < 0 , 2 > = < √3 , 3 >

Use the Law of Cosines to find angle θ between the given vectors (0° ≤ θ ≤ 180° )
Example 4:
v = 3i + j and w = 2i - j
v = < 3 , 1 > and w = < 2 , -1 >

v + w = < 3 + 2 , 1 + -1 > = < 5 , 0 > so
the ll v + w ll = √(5 2 + 02) = 5

ll v ll = √( 3 2 + 12) = √(9 + 1) = √(10)
ll w ll = √( 2 2 + (-1)2) = √(4 + 1) = √(5)
ll v + w ll = 5

Using the Law of Cosines:

52 = (√(10)) 2 + (√(5))2 - 2 (√(10))(√(5))Cos θ

25 = 10 + 5 - 2 √(50) Cos θ

10 = - 2 √(50) Cos θ

-.7071067812 = Cos θ

135 ° = θ

Example 5: Find the angle between the forces given the magnitude of their resultant:
Force 1 = 3000 pounds
Force 2 = 1000 pounds
Resultant Force = 3750 pounds

37502 = 10002 + 30002 - 2(1000)(3000)Cos β

-.67708333333 = Cos β

β = 132.6161427 °

θ = 47.38385729°

Monday, March 19, 2007

6.3 Vectors in the Plane

6.3 Vectors in the Plane

Check out the following web sites:
http://www.math.duke.edu/education/ccp/materials/mvcalc/vectors/vec1.html

Vectors are used to analyze numerous aspects of everyday life.
Vector in the plane - given vector PQ written bold faced v = PQ (with a partial ray over the PQ but with only the top part of the arrow and with the letters PQ under it)
Vectors are denoted by lowercase, boldface letters such as u, v, and w.

Vectors are a directed line segment vector PQ having initial point P and terminal point Q.
It has magnitude, or length, is denoted by ll PQ ll.

If the directed line segment whole initial point is the origin, then the representative of the vector v is in standard position. The component form of a vector v, written v = <v1, v2>

otherwise:

If the directed line segment initial oint is P and the terminal point is Q then
P = (p
1, p2) and Q = (q1, q2) then vector PQ = <q1 - p1, q2 - p2> = <v1, v2> = v

If both the initial point and the terminal point lie at the origin, v is the zero vector and denoted by 0 =
<0, 0>

Example 1: Given the initial point P (1 , 0) and the terminal point Q (3 , 4) give the component form of vector PQ

P = ( p1, p2) and Q = (q1, q2)
vector PQ =
< q1 - p1 , q2 - p2> = <v1, v2> = v = < 3 - 1 , 4 - 0 > = < 2, 4 >

Magnitude or length of v is

ll vll =
√((q1 - p1)2 + (q2 - p2)2 ) = √(v12, v22)
If llvll = 1, v is a unit vector.

llvll = 0 if and only if v is the zero vector 0.

llvll =
√((3 - 1)2 + (4 - 0)2 ) = √(22, 42) = √(4 + 16) = √20
= √4√5 = 2√5

Parallelogram Law - for vector addition is the sum of vector v and vector u, v + u, is often called the resultant of vector addition, is the diagonal of a parallelogram having u and v as its adjacent sides

Definition of Vector Addition and Scalar Multiplication
Let u =
<u1, u2> and v = <v1, v2>
be vectors and let k be a scalar (a real number).

Then the sum of u and v is the vector
u + v =
<u1+ v1 , u2 + v2>

and the scalar multiple of k time u is the vector
ku = k
<u1, u2> = <ku1, ku2>

Properties of Vector Addition and Scalar Multiplication:

Let u, v, and w be vectors and let “c” and “d” be scalars. Then the following properties are true.

1. u + v = v + u Addition commutative property

2. (u + v) + w = u + (v + w) Addition associative property

3. u + 0 = 0 zero property

4. u + (-u) = 0 Additive inverse

5. c (du) = (cd) u Multiplication associative property

6. (c + d ) u = cu + du Distributive property

7. c (u + v) = cu + cv Distributive property

8. 1 (u) = u, 0 (u) = 0

9. llcvll = lcl llvll

u = unit vector = (v)divided by the magnitude of v
The vector u is called a unit vector in the direction of v.

Example 2: let u = <0, -9> and v = <-6, 10>,

a). find u + v

<0, -9> + <-6, 10> = <0 + -6, -9 + 10> = <-6, 1>

b). find u - v

<0, -9> - <-6, 10> = <0 - -6, -9 - 10> = <6, -19>

c). find 2u - 3v

2 <0, -9> - 3<-6, 10> = <0, -18> + <18, -30> = <0 + 18, -18 - 30> = <18, -48>

d). find v + 4u

<-6, 10> + 4<0, -9> = <-6, 10> + <0 , -36> = <-6 + 0, 10 + -36> = <-6, -26>

Standard Unit Vectors
i =
<1, 0> and j = <0, 1>

Given
v = <v1, v2>
v = v1 <1, 0> + v2 <1, 0>
= v1 i + v2 j

is the Linear Combination of the vectors i and j

Writing a Linear Combination of Unit Vectors
Example 3: Given initial point P (5, -12) and terminal point Q (-4, 7)

Begin by writing the component form of the vector PQ

<5 + -4, -12 + 7> = <1, -5>
Then to write vector PQ as a linear combination of the standard unit vectors i and j
vector PQ = 1i -5j

Example 4: let u = 2i - j and v = -i + j
u + v = 2i - j + -i + j = i so therefore u + v = <1, 0>

to check, recall i = <1, 0> and j = <0, 1> so therefore

u = 2 <1, 0> - <0, 1> = <2, 0> - <0, 1> = <2, -1>
v = - <1, 0> + <0, 1> = <-1, 0> + <0, 1> = <-1, 1>
so u + v = <2, -1> + <-1, 1> = <2 + -1, -1+ 1> = <1, 0>

Example 5: Vector Operations:
let u = -2i + 3j and v = 4i + 5j
Find 2u - 3v

2(-2i + 3j) - 3(4i + 5j) = -4i + 6j - 12i -15j
= -4i -12i + 6j -15j = -16i -9j

Tuesday, March 13, 2007

Precalculus 9.5 The Binomial Theorem

9.5 The Binomial Theorem

Binomial is a polynomial that has 2 terms

(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 +2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5 = 1x5 + 5x4y + 10 x3y2 + 10 x2y3 + 5xy4 + 1y5
Notice the pattern that is developing in each expansion of the binomial.
1. The sum of the exponents of each term is “n”.
2. In each expression, there are n + 1 terms.
3. In each expansion, x and y have symmetric roles. The powers of x decreases by 1 in successive terms, whereas the powers of y increase by 1.
4. The coefficient increase and decrease in a symmetric pattern.

The coefficients of a binomial expansion are called binomial coefficients.
A display that contains only the coefficients of the terms in the expansions is called Pascal’s Triangle. The first triangle is Pascal’s Triangle, the second is using combinations to find the numbers.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

0C0
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
5C0 5C1 5C2 5C3 5C4 5C5

Therefore in general, The Binomial Theorem: In the expansion of

(x + y)n = nC0 (x)n y0 + nC1 (x)n-1 y1 + nC2 (x)n-2 y2 + ... + nCn-1 (x)1 yn-1 + nCn (x)0 yn

The coefficient of xn-r yr is
The symbol ( ) is often used in place of nCrto denote binomial coefficients.

Example 1: Find the following pairs of binomial coefficients.

a. 7C0 , 7C7 = 1 , 1
b. 7C1 , 7C6= 7 , 7
c. 8C2 , 8C6 = 28 , 28 same answer

What can you conclude?

nCr = nCn - r
This shows the symmetric property of binomial coefficients.

(2x - 3y)3 = 3C0 (2x)3(-3y)0 + 3C1 (2x)2 (-3y)1 + 3C2(2x)1(-3y)2 + 3C3 (2x)0(-3y)3

= (1)(8x3)(1) + (3)(4x2)(-3y) + (3)(2x)(9y2) + (1)(1)(-27y3)

= 8x3 - 36x2y + 54xy2 - 27y3

Remember to put the negative sign in the parenthesis so when you raise it to a power, the answer is appropriately positive or negative!!

Expand (2x - y)5 = (2x)5 + 5(2x)4(-y) + 10(2x)3(-y)2 + 10(2x)2(-y)3 + 5(2x)(-y)4 + (-y)5

= 32x5 - 80x4y + 80 x3y2 - 40 x2y3 + 10xy4- y5

Do you notice a pattern for (x - y)n ?

Precalculus 9.4 Mathematical Induction

9.4 Mathematical Induction
- a form of mathematical proof

Lets look at the patterns:

S4 = 1 + 2 + 3 + 4 = 10
S6 = 1 + 2 + 3 + 4 + 5 + 6 = 21
S8 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
S10 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

When n = 4, S4 = 10, what are the factors of 10? (2)(5)
When n = 6, S6 = 21, what are the factors of 21? (3)(7)
When n = 8, S8 = 36, what are the factors of 36 that fit the pattern? (4)(9)
When n = 10, S10 = 55, and again, the factors are (5)(11)
therefore we can conclude that Sn = (n/2)(n + 1)

Now lets try another one:
S1 = 1 = 12
S2 = 1 + 3 = 22
S3 = 1 + 3 + 5 = 32
S4 = 1 + 3 + 5 + 7 = 42
S5 = 1 + 3 + 5 + 7 + 9 = 52

Sn = 1 + 3 + 5 + 7 + ... + (2n - 1) = n2

Looking at the pattern we see that 2n - 1 represents consecutive odd integers and n2 represents the sum of the odd integers.

By looking at patterns, sometimes we may come up with a wrong conclusion:
Example:
Take a circle and put a dot on it. How many areas do you have? 1
Now put 2 dots on it and connect the dots. How many areas do you have? 2
Now put 3 total dots on it and connect the dots. How many areas do you have? 4
Now put 4 total dots on it and connect the dots. How many areas do you have? 8
Now put 5 total dots on it and connect the dots. How many areas do you have? 16

Lets look at what we have seen -
1 to 1
2 to 2
3 to 4
4 to 8
5 to 16
so what do you conclude 6 would go to?
Normally we would say 32, but it actually has 30 or 31 depending if the dots are equidistant from each other. So this pattern does not work.

Therefore we have to prove a formula works for all numbers, you have to do the following.

The Principle of Mathematical Induction:

Let Pn be a statement involving the positive integer n. If:
1. P1 is true AND
2. The truth of Pk implies the truth of Pk+1 , for every positive k, then Pn must be true for all positive integers n.

Example:

So Sn = 1 + 3 + 5 + 7 + ... + (2n - 1) = n2
Let n = 1

1. S1 = 2(1) - 1 = 1 and 12 = 1 both are true

2. Then Assume Sk = 1 + 3 + 5 + 7 + ... + (2k - 1) = k2

Then Sk+1 = 1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1 ) = (k + 1)2

Sk+1 = Sk + (2k + 2 - 1 ) = (k + 1)2

k2 + 2k + 1 = (k + 1)2
(k + 1)2 = (k + 1)2

this is what we had to prove so
Therefore this formula is valid for all positive integer values for n.

Example 2: Find Pk+1 for the given Pk
Pk = (k/2)(5k - 3)
Pk+1 = ((k + 1)/2)(5(k+1) - 3)
= ((k + 1)/2)(5k + 5 - 3)
= ((k + 1)/2)(5k + 2)
= (1/2)(k + 1)(5k + 2)
= (1/2)(5k2 + 7k + 2)

Example 3:
Sn = 1 + 5 + 9 + 13 + ... + (4n - 3) = n (2n - 1)

Let n = 1
S1 = 4(1) - 3 = 1 and (1)(2(1) - 1) = (1)(1) = 1 therefore it is true for both

Assume:
Sk = 1 + 5 + 9 + 13 + ... + (4k - 3) = k (2k - 1)

then Sk+1 = 1 + 5 + 9 + 13 + ... + (4k - 3) + (4(k + 1) - 3) = (k + 1)(2 (k + 1) - 1)
Sk+1 = Sk + (4k + 4 - 3) = (k + 1)(2 k + 2 - 1)
Sk+1 = k (2k - 1) + (4k + 1) = (k + 1)(2 k + 1)
Sk+1 = 2k2 - k + 4k + 1 = (k + 1)(2 k + 1)
Sk+1 = 2k2 + 3k + 1 = (k + 1)(2 k + 1)
Sk+1 =(k + 1)(2 k + 1) = (k + 1)(2 k + 1)

Therefore the formula is valid for all positive integer values of n.

Example 4: 2 (1 + 3 + 32 + 33 + ... + 3n -1) = 3n - 1
Sn = 2 (1 + 3 + 32 + 33 + ... + 3n -1) = 3n - 1

Let n = 1

S1 = 2 (31-1) = 2(30) = 2(1) = 2 and 31 - 1 = 3 - 1 = 2

Assume Sk = 2 (1 + 3 + 32 + 33 + ... + 3k -1) = 3k - 1

Sk+1 = 2 (1 + 3 + 32 + 33 + ... + 3k -1 + 3(k+1)-1 )= 3k+1 - 1
Sk+1 = 2 (1 + 3 + 32 + 33 + ... + 3k -1 + 3k ) = 3k+1 - 1
Sk+1 = 2 (1 + 3 + 32 + 33 + ... + 3k -1 ) + 2(3k ) = 3k+1 - 1
Sk+1 = Sk + 2(3k ) = 3k+1 - 1
Sk+1 = 3k - 1 + 2(3k ) = 3k+1 - 1
Sk+1 = 3(3k ) - 1 = 3k+1 - 1
Sk+1 = 3k+1 - 1 = 3k+1 - 1

We can conclude that this formula holds for all positive integer values of n.

Sums of Powers of Integers - find on
http://www.libraryofmath.com/summation-formulas.html

or in textbook page 652

a0 = 3, a1 = 3, a4 = 15, an = an2 + bn + c

a0 = a(0)2 + b(0) + c = 3 therefore c = 3

a1 = a(1)2 + b(1) + c = 3
= a + b + 3 = 3
= a + b = 0
a = -b

a4 = a(4)2 + b(4) + c = 15
= 16a + 4b + 3 = 15
= 16a + 4b = 12
16(-b) + 4b = 12
-12b = 12
b = -1

a = - b
a = -(-1) = 1

therefore an = n2 - n + 3

To check your answer, use the TI-83 calculator, put the data in:
L1 (0, 1, 4) and L2 (3, 3, 15)
Stat go to Quad Reg L1, L2

this gives you an = n2 - n + 3

Monday, March 5, 2007

9.3 Geometric Sequences and Series

9.3 Geometric Sequences and Series

Geometric Sequences - consecutive terms have a common ratio.

A. Definition of Geometric Sequences:

A sequence is a geometric sequence if the ratios of consecutive terms are the same.

(a2/a1) = (a3/a2) = (a4/a3) = ... = r, r cannot equal 0

The number r is the common ratio of the sequence.

Example 1: given 1, -2, 4, -8, ... , find the common ratio

(-2/1) = (4/ -2) = (-8/ 4) = -2

Example 2: write the first five terms of the geometric sequence.
a1 = 10 , r = 2

(a2/a1) = r = 2
(a2/10) = 2
a2 = 20

(a3/a2) = 2
(a3/20) = 2
a3 = 40

(a4/a3) = 2
(a4/40) = 2
a4 = 80

(a5/a4) = 2
(a5/80) = 2
a5 = 160

B. The nth Term of a Geometric Sequence:
the nth term of a geometric sequence has the form

an = a1 rn-1

where r is the common ratio of consecutive terms of the sequence. So, every geometric sequence can be written in the following form.
a1, a2, a3, a4, a5... , an, ... which is the same as

a1, a1r, a1r2, a1 r3, a1 r4 , ..., a1 rn-1, ...

Therefore
a1 = a1
a2 = a1r
a3 = a1r2
a4 = a1r3
a5 = a1r4
an, = a1 rn-1

Example 3: write the first five terms of the geometric sequence. Determine the common ratio and write the nth term of the sequence as a function of n.

a1 = 81, ak+1 = (1/3) ak
a2 = a1+1 = (1/3) a1 = (1/3) (81) = 27
a3 = a2+1 = (1/3) a2 = (1/3) (27) = 9
a4 = a3+1 = (1/3) a3 = (1/3) (9) = 3
a5 = a4+1 = (1/3) a4 = (1/3) (3) = 1

Therefore the common ratio is (27/81) = (9/27) = (3/9) = (1/3 )
an = a1 rn-1 = 81 (1/3)n-1 = (1/3)-4 (1/3)n-1 = (1/3)-4 +n-1 = (1/3)n-5

Example 4: Find the 7th term of the following geometric sequence: 3, 36, 432

r = (36/3) = 432/36) = 12
a1 = 3
a7 = a1r7-1 = 3 (12)6 = 3 (2985984) = 8957952

C. The Sum of a Finite Geometric Sequence

The sum of the geometric sequence
a1, a1r, a1r2, a1 r3, a1 r4 , ..., a1 rn-1
with the common ratio r is not equal to 1 is

Sn = a1 ((1 - rn)/(1 - r))

Example 5: Finding the sum of a finite geometric sequence

(-2)1-1 + (-2)2-1 + (-2)3-1 + (-2)4-1 + (-2)5-1 + (-2)6-1 + (-2)7-1 + (-2)8-1 + (-2)9-1 =
(-2)0 + (-2)1 + (-2)2 + (-2)3 + (-2)4 + (-2)5 + (-2)6 + (-2)7 + (-2)8 =
1 + -2 + 4 + -8 + 16 + -32 + 64 + -128 + 256 = 171
or
Sn = a1 ((1 - rn)/(1 - r))
S9 = a1 ((1 - r9)/(1 - r)) = 1(1 - (-2)9)/(1 –-2) = (1 - -512) / (3) = (513) / (3) = 171

C. The Sum of a Infinite Geometric Sequence

then the infinite geometric series

a1, a1r, a1r2, a1 r3, a1 r4 , ..., a1 rn-1 , ...

has the sum

S = a1 / (1 - r)

Example 6: Find the sum of the infinite geometric series

recall: an = a1 rn-1 so therefore a1= 1 and r = 2

Sn = (1- rn) / (1 - r)
since rn = 2 = ∞
Sn = (1- ∞ ) / (1-2) = -∞ / (- 1) = ∞

Example 7: Find the sum of the infinite geometric series

S = a1 / (1 - r)
Since the first term n=0 then 4(.2)0 = 4 therefore

= 4 / (1 - .2) = 4 / .8 = 5

9.3 Homework #54; page 645; # 11 - 23 odd, 31 - 45 odd, 55 - 83 odd
Quiz on Sections 9.1 - 9.3 next class

9.2 Arithmetic Sequences and Partial Sums

9.2 Arithmetic Sequences and Partial Sums

Arithmetic Sequences - a sequence whose consecutive terms have a common difference

A. Definition of Arithmetic Sequences:
A sequence is arithmetic if the differences between consecutive terms are the same. So, the sequence

a1, a2, a3, a4, ... , an, ...

is arithmetic if there is a number d such that

a2 - a1 = a3 - a2 = a4 - a3 = ... = d.

The number d is the common difference of the arithmetic sequence.

Example 1: Given the sequence an = 1 + (n - 1)4, find the first five terms of the sequence an the common difference:
a2 = 1 + (2 - 1)4 = 1 + (1)(4) = 1 + 4 = 5
a3 = 1 + (3 - 1)4 = 1 + (2)(4) = 1 + 8 = 9
a4 = 1 + (4 - 1)4 = 1 + (3)(4) = 1 + 12 = 13
a5 = 1 + (5 - 1)4 = 1 + (5)(4) = 1 + 20 = 21

a2 - a1 = a3 - a2 = a4 - a3 = ... = d.

9 - 5 = 13 - 9 = 21 - 13 = 4 = d

B. The nth term of an Arithmetic Sequence: last year you learned it as

an = a1 + (n - 1) d
distributing d we get: an = a1 + dn - d = dn + (a1 - d) therefore let’s let c = (a1 - d) so we have

an = dn + c

where d is the common difference between consecutive terms of the sequence and c = a1 - d

Example 2: write the first five terms of the arithmetic sequence, find the common difference and write the nth term of the sequence as a function of n.

a1 = 200, ak+1 = ak - 20
a2 = a1+1 = a1 - 20 = 200 - 20 = 180
a3 = a2+1 = a2 - 20 = 180 - 20 = 160
a4 = a3+1 = a3 - 20 = 160 - 20 = 140
a5 = a4+1 = a4 - 20 = 140 - 20 = 120
so d = -20 and c = a1 - d = 200 - -20 = 220
therefore an = dn + c = -20n + 220

Example 3: The first two terms of the arithmetic sequences are given. Find the missing term.
a1 = 3, a2 = 13 , a9 = ____

d = 13 - 3 = 10
c = a1 - d = 3 - 10 = -7
so therefore an = dn + c = 10n - 7

a9 = 10n - 7 = 10 (9) - 7 = 90 - 7 = 83

Example 4: Find a formula for an for the arithmetic sequence:
a1 = 15, d = 4

c = a1 - d = 15 - 4= 11
so therefore an = dn + c = 4n +11

Example 5: Find a formula for an for the arithmetic sequence:
a1 = -4, a5 = 16

You may have noticed by now that an = dn + c is a linear function so therefore d is the slope of the line so using this concept of the slope formula ▵y/▵x We have:

(16 - -4) / (5 - 1) = (20)/ (4) = 5 = d

c = a1 - d = - 4 - 5 = -9
so therefore an = dn + c = 5n - 9

C. The Sum of a Finite Arithmetic Sequence

Carl Friedrich Gauss (1777 - 1855) was ten years old, his teacher asked him to add all the integers from 1 to 100. Gauss was able to answer the teacher within a few moments, the teacher was amazed. Here is how he did it:

1 + 2 + 3 + 4 + ... + 97 + 98 + 99 + 100

He took the 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101 so taking 100 / 2 = 50, then 50 * 101 = 5050.

The Sum of a Finite Arithmetic Sequence:
The sum of a finite arithmetic sequence with n terms is

Sn = (n/2)(a1 + an)

Example 6: find the indicated nth partial sum of the arithmetic sequence:

-6, -2, 2, 6, ... , n = 50

the common difference is -2 - -6 = 2 - -2 = 6 - 2 = 4
c = a1 - d = - 6 - 4 = -10
so therefore an = dn + c = 4n - 10

a1 = -6 and a50 = 4(50) - 10 = 200 - 10 = 190
Sn = (n/2)(a1 + an) = (n/2)(a1 + a50) = (50 / 2) (-6 + 190) = (25)(184) = 4600

9.2 Homework #53 page 635; # 9 - 19 odd, 35 - 47 odd, 59 - 79 odd

Precalculus 9.1 Sequences and Series

9.1 Sequences and Series

A sequence is a function whose domain is the set of positive integers.

Definition of Sequence:
An infinite sequence is a function whose domain is the set of positive integers. The function values
a1, a2, a3, a4, ... , an, ...
are the terms of the sequence. If the domain of the function consists of the first n positive integers only, the sequence is a finite sequence.

A. Finding the Terms of a Sequence:
Example 1: find the first five terms of an = 4n - 7
a1 = 4(1) - 7 = -3
a2 = 4(2) - 7 = 8 - 7 = 1
a3,= 4(3) - 7 = 12 - 7 = 5
a4 = 4(4) - 7 = 16 - 7 = 9
a5 = 4(5) - 7 = 20 - 7 = 13

Example 2: Find the 16th term of the sequence an = (-1)n-1(n(n-1))
a16 = (-1)16-1(16(16-1)) = (-1)15(16(15)) = (-1)(240) = -240

Example 3: Write the first five terms of the sequence defined recursively:
a1 = 15, ak+1 = ak+ 3
Let k = 1 so we have:
a1+1 = a2 = a1 + 3 = 15 + 3 = 18
Let k = 2 so we have:
a2+1 = a3 = a2 + 3 = 18 + 3 = 21
a3+1 = a4 = a3 + 3 = 21 + 3 = 24
a4+1 = a5 = a4 + 3 = 24 + 3 = 27

B. Finding the nth term of a Sequence
Example 4: Write an expression for the apparent nth term of the sequence
(assume n begins with 1): 3, 7, 11, 15, 19, ...

7 - 3 = 11 - 7 = 15 - 11 = 4

As you can see, the terms are going up by 4 and the first term is one less than 4 so
an = 4n - 1

Example 5: 1, 1/4, 1/9, 1/16, 1/25, ...
As you can see, the terms denominators are perfect squares so
an = 1/(n2) = n-2

C. The Fibonacci Sequence: A Recursive Sequence
The Fibonacci sequence is defined recursively as follows.
a0 = 1, a1 = 1, ak = ak-2 + ak-1 , where k is greater than or equal to 2
{1, 1, 2, 3, 5, 8, ...}

Example 6: Write the first five terms of the sequence defined recursively. Use this pattern to write the nth term of the sequence as a function of n.
a1 = 25, ak+1 = ak - 5
a2 = a1+1 = a1 -5 = 25 -5 = 20
a3 = a2+1 = a2 -5 = 20 - 5 = 15
a4 = a3+1 = a3 -5 = 15 - 5 = 10
a5 = a4+1 = a4 -5 = 10 - 5 = 5
Therefore an = 25 - 5n

C. Definition of Factorial:
In n is a positive integer, n factorial is defined by
n! = 1 x 2 x 3 x 4 x ... (n - 1) x n
As a special case, zero factorial is defined as 0! = 1

Example 7:
Simplify the ratio of factorials.
(4!)/(7!) = (4!) / (7 x 6 x 5 x 4!) = 1/ (7 x 6 x 5 ) = 1 / 210

Example 8: Simplify the ratio of factorials.
(n + 2)! / (n!) = (n + 2)(n + 1)(n!) / (n!) = (n + 2)(n + 1) = n2 + 3n + 2

D. Definition of Summation Notation:
The sum of the first n terms of a sequence is represented by

The summation of ai when i = 1 to i = n is equal to

a1 + a2 + a3 + a4 + ... + an

where i is called the index of summation, n is the upper limit of summation, and 1 is the lower limit of summation. (check out http://www.cs.fsu.edu/~cop4531/slideshow/chapter3/3-1.html)

Example 9: Find the sum of 3i - 1 when i = 1 to i = 6

3 - 1 + 6 - 1 + 9 - 1 + 12 - 1 + 15 - 1 + 18 - 1 = 57

To use your TI - 83, TI - 83 plus calculators:
Go to mode - change from function to sequential mode, then
2nd Stat arrow to math #5 Sum, 2nd Stat arrow to ops #5 seq then put in the sequence, then n to show the calculator the variable, then the lower bound number, then the upper bound number, then close the parenthesis twice, then press enter.
Your screen should look like this:
sum(seq(3n-1,n,1,6)) = 57

E. Properties of Sums - http://www.libraryofmath.com/summation-formulas.html

F. Series:

Definition of a Series:
Consider the infinite sequence a1, a2, a3, a4, ... , ai, ...
1. The sum of all terms of the infinite sequence is called an infinite series and is denoted by
a1 + a2 + a3 + a4 + ... + ai + ... = The summation of ai when i = 1 to i = infinity.

2. The sum of the first n terms of the sequence is called a finite series or the nth partial sum of the sequence and is denoted by
a1 + a2 + a3 + a4 + ... + an = The summation of ai when i = 1 to i = n.

Example 10: Find the sum of the partial sum of the series:

The summation of 8(-1/2)n when n = 1 to n = infinity, the 4th partial sum

= 8(-½ )1 + 8(-½ )2 + 8(-½ )3 + 8(-½ )4 = -4 + 2 + -1 + .5 = -5/2

(Calculator check: you can check your individual answers by putting you calculator in function mode, put the series into y1 = 8 (-.5)x and looking in your table to get your values.)

Example 11
: Find the sum of the infinite series:

The summation of 4(1/10)n when n = 1 to n = infinity

= 4 [ .1 + .01 + .001 + .0001 + ...] = 4 (.111111...) = 4 (1/9) = 4/9

Example 12: Find the sum of the infinite series:

The summation of 8(1/10)n when n = 1 to n = infinity is equal to

= 8 [ .1 + .01 + .001 + .0001 + ...] = 8 (.111111...) = 8 (1/9) = 8/9