Sunday, March 13, 2011
CH 9 Sequences and Series
Chapter 3 Review
a- causesthe graph to vertically stretch or shrink and changes y-intercept
d- causes the graph to move up and down
a^x / a^y = a^(x-y)
a^-x = 1 / a^x
f(x) = e^x
Domain: (-∞, ∞)
Range: (0, ∞)
y-intercept: (0,1)
Continuously Compounded Interest: A = Pe^rt
A = present value
P = principle value (how much invested)
r = annual rate of return (decimal form)
t = time in years
n = number of compounding periods per year
x-intercept: (0,1)
Thursday, March 10, 2011
Chapter 5 Review
Reciprocal
csc(x)=1/sin(x)
Quotient
tan(x)=sin(x)/cos(x)
Pythagorean
sin^2(x)+cos^2(x)=1
Cofunction
sin((pi/2)-x)=cos(x)
Even/Odd
cot(-x)=-cot(x)
5.2 goes over verifying identities. Some guidelines are:
- Work with one side of the equation at a time, preferably the more complicated side.
- Look for opportunities to factor an expression, add fractions, square a binomial, or create a common denominator.
- Look for opportunities to use the fundamental identities. Note which functions are in the final expression you want.
- Try converting terms to sines and cosines.
For help with sections 5.1 and 5.2: http://www.youtube.com/watch?v=pviWtesNnAY&feature=related
5.3 covers solving trigonometric equations. When solving an equation try:
- Combining like terms
- Taking squares and square roots
- Factoring
- Rewriting trigonometric functions
- Using inverse functions
5.4 goes over the sum and difference formulas. These are used in identities and to evaluate values that aren't on the unit circle. Here are some examples:
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
5.5 covers multiple-angle and product-sum formulas. An example of a double-angle formula is:
- sin(2x)=2sin(x)cos(x)
An example of a power-reducing formula is:
- cos^2(x)=(1+cos(2x))/2
An example of a half-angle formula is:
- tan(x/2)=(1-cos(x))/sin(x)=sin(x)/(1+cos(x))
An example of a product-to-sum formula is:
- sin(x)cos(y)=1/2(sin(x+y)+sin(x-y))
And an example of a sum-to-product formula is:
- cos(x)+cos(y)=2cos((x+y)/2)cos((x-y)/2)
Good luck everyone!
Limits: Review
If f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, the limit of f(x) as x approaches c is L. This is written as
Lim f(x) = L
x ->c
(x approaches c, as y approaches L)
Eample:
f(x) = 2x-3
Lim f(x)
x->4
2(4) - 3
8 - 3 = 5
Lim f(x) = 5
X->4
Conditions Under Which Limits Do Not Exist:
(The limit of f(x) as x-> c does not exist if any of the following conditions is true.
1. f(x) approaches a different number from the right side of c than from the left side of c
2. f(x) increases or decreases without bound as x approaches c
3. f(x) oscillates between two fixed values as x approaches c
Properties of Limits:
(Let b and c be real numbers and let n be a positive integer)
lim b = b
x-> c
lim x = c
x->c
lim x^n = c^n
x->c
Techniques for Evaluating Limits
Evaluating Limits by Direct Substitution:
Lim (x^2 + x – 6)
x-> -1
substitute -1 in for x
(-1)^2 + (-1) – 6 = -6
Dividing out Technique:
Lim (x^2 + x – 6/x +3)
Begin by factoring the numerator and dividing out any common factors
Lim (x^2 + x – 6/x +3) = lim (x -2)(x +3)/(x + 3)
= lim (x -2)
= ((-3) – 2)
= -5
This procedure for evaluating a limit is called the dividing out technique. The validity of the procedure stems from the fact that if two functions agree at all but a single number c, they must have identical limit behavior at x = c.
Rationalizing Technique:
In this case, we change the form of the fraction by rationalizing (eliminating the radical in) the numerator:
lim f(x) = lim f(x) = f(c)
If f is a function and L1 and L2 are real numbers, the statements
Lim f(x) = L1 limit as x approaches negative infinite
X->infinite
Lim f(x) = L2 limit as x approaches infinite
x-> infinite
Limits at Infinity:
If r is a positive real integer, then
Lim 1/x^r = 0
x-> infinite
Furthermore, if x^r is defined when x<0, r =" 0"> - infinite
Comparing Limits at Infinity:
Lim -2x + 3/3x^2 = 1
x->infinite
= 0
x-> infinite
= -2/3
x-> infinite
= DNE
Friday, February 18, 2011
Evaluating Limits
Wednesday, February 16, 2011
12.1 Continued
There are 4 properties:
Lim b = b
x→ c
Lim x = c
x→ c
Lim xn = cn
x→ c
Ex:
(1st property)
Lim 5 = 5
x→2
(2nd property)
Lim x = 2
x→3
(3rd property)
Lim x2 = 9
x→3
(4th property)
Lim (cubed root x) = 2
x→8
Operations with Limits
Lim f(x) = L
x→c
and
Lim g(x) = K
x→c
Scalar Multiple:
Lim [bf(x)]= bL
x→c
Sum or difference:
Lim [f(x) ± g(x)] = L ± K
x→c
Product:
Lim [f(x)g(x0] = LK
x→c
Quotient:
Lim [f(x)/g(x)] = L/K
x→c
****with this one K can not be 0
Power:
Lim [F(x)]n = Ln
x→c
Continuity
Function is continuous if:
Lim f(x) = f(a)
x→a
For extra help with continuity
http://www.youtube.com/watch?v=hlorAjS0xWE&feature=related
Introduction to Limits
Wednesday, February 9, 2011
Binomial Theorem
Sunday, February 6, 2011
Geometric Sequences and Series
The sequence whose nth term is 2^n is geometric. For this seqence, the common ratio between consecutive terms is 2.
if r is less that absolute value 1 then the equation is s=a1/1-r
Need more help, follow the link
soft
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Thursday, February 3, 2011
Arithmetic Sequences and Partial Sums
Sn = (n/2)(a1 + an)
You can use this equation to add up the numbers in a Arithmetic Sequence.
Example:Sn = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 +.....+ 98 + 99 + 100
= (100/2)(1 + 100)
= 50(101)
=5050
You can find a Partial Sum of an Arithmetic Sequence by using the equation for the Sum of a Finite Arithmetic Sequence.
Example:
5, 16, 27, 38, 49,....
Find the 150th term
an = dn + c
a150 = 11(150) - 6
a150 = 1644
S150 = (150/2)(5 + 1644)
S150 = 75(1649)
S150 = 123,675
Sunday, January 30, 2011
Sequence and Series
Finding the Terms of a sequence
The Fibonacci Sequence: A Recursive Sequence
- 0!=1
- 1!=1
- 2!=1 x 2 = 2
- 3!= 1 x 2 x 3 = 6
Evaluating Factorial Expressions
Definition of Summation Notation- The sum of the first n terms of a sequence is represented by
Where i is called the index of summation, n is the upper limit of summation, and 1 is lower the limit of summation
Sigma Notation for Sums