Math

Sequence

A list of numbers with a pattern.

Series

The sum of numbers in a sequence. Take that list of numbers separated by commas and replace it with a +.

Infinite Sequence

Sequences that have infinite terms

Common Log

y=log₁₀x

Natural Log

y=lnx OR y=logₑx

Property of Logarithm

Partial Fractions

1. Check the degree (highest exponent): The degree of the numerator must be strictly less than the degree of the denominator. If it isn't (e.g., \(x^{3}\) on bottom and \(x^{2}\) on top), you must use polynomial long division first to simplify it.

2. Factor the denominator: Break the denominator down into its simplest multiplied components (factors).

3. Set up the fraction template: Assign unknown constants (usually \(A, B, C\), etc.) for the numerators over each factor.

4. Solve for the constants: Multiply the entire equation by the common denominator to eliminate the fractions, then equate the coefficients or plug in smart values for \(x\) to solve for $A, B$, and \(C\).

• Repeated Linear Factors: If a factor repeats, like \((x-a)^2\), you must account for each power: \(\frac{A}{x-a} + \frac{B}{(x-a)^2}\).

1. For each irreducible quadratic factor 𝑎⁢𝑥2+𝑏⁢𝑥+𝑐 that 𝑄⁡(𝑥) contains, the decomposition must include
   

   𝐴⁢𝑥+𝐵𝑎⁢𝑥2+𝑏⁢𝑥+𝑐.

2. For each repeated irreducible quadratic factor (𝑎⁢𝑥2+𝑏⁢𝑥+𝑐)𝑛, the decomposition must include
   

   𝐴1⁢𝑥+𝐵1𝑎⁢𝑥2+𝑏⁢𝑥+𝑐+𝐴2⁢𝑥+𝐵2(𝑎⁢𝑥2+𝑏⁢𝑥+𝑐)2+⋯+𝐴𝑛⁢𝑥+𝐵𝑛(𝑎⁢𝑥2+𝑏⁢𝑥+𝑐)𝑛.

Ex: ∫x²+3x+1/(x+2)(x-3)²(x²+4)² → A/x+2 + B/(x-3)^1 + C/(x-3)² + Dx+E/(x²+4)^1 + Fx+G/(x²+4)²

Double angle identity for sin2x

2sinxcosx

Double angle identity for cos2x

cos²x-sin²x, 2cos²x-1, 1-2sin²x

Double angle identity for tan2x

2tanx/1-tan²x

Power Reduction for sin²x

1-cos2x/2

Power Reduction for cos²x

1+cos2x/2

Power Reduction for tan²x

1-cos2x/1+cos2x

Power Reduction for csc²x

2/1-cos2x

Power Reduction for sec²x

2/1+cos2x

Power Reduction for cot²x

1+cos2x/1-cos2x

Product to Sum Formula for sinAcosB

(1/2) [sin(A+B) + sin(A-B)]

Product to Sum Formula for cosAsinB

(1/2) [sin(A+B) - sin(A-B)]

Product to Sum Formula for cosAcosB

(1/2) [cos(A+B) + cos(A-B)]

Product to Sum Formula for sinAsinB

(1/2) [cos(A-B) - cos(A+B)]

Cofunction identities

When you integrate a product of cosine and sin, where the exponent for one of the functions is 1, and the other is >1

let u=the trig function with the higher power

If you have the integral of cos²x sin³x dx

Take the one with the odd exponent and split it up, and then use pythag identity on what you just split up

-cosx + C

∫sinx dx

sinx + C

∫cosx dx

tanx + C

∫sec²x dx

-cotx + C

∫csc²x dx

secx + C

∫secx tanx dx

-cscx + C

∫cscx cotx dx

ln|secx + tanx| +C

∫secx dx

ln|secx| +C

∫tanx dx

Derivatives of Inverse Trig Functions

PSST Rule

Sec → Sec ← Tan
Derivative of secx = secxtanx
Integral of secxtanx = secx

Derivative of tanx = sec²x
Integral of sec²x = tanx

CCC Rule

Csc → -Csc ← Cot
Derivative of cscx = -cscxcotx
Integral of -cscxcotx = cscx

Derivative of cotx = -csc²x
Integral of -csc²x = cotx

Derivate of ln|x|

= 1/x

Integral of ln(x)

xln(x) - x + C

Product Rule for Log

logₙ(xy)=logₙx+logₙy

Quotient Rule for Log

logₙ(x/y) = logₙx-logₙy

Power Rule for Log

logₙ(x^b) = b logₙx

Change of Base Rule for Log

logₙx = log(sub c) x / log (sub x)b

Zero Rule for Log

logₙb(1) = 0

Pythagorean Identities

csc²x + sec²x = 1

sin²x + cos²x = 1

tan²x + 1 = sec²x

Arithmetic Sequences

add a d(difference) to each term

formula: a₁+(n-1)d

find d: a(sub)n+1 - a(sub)n

Geometric Sequence

multiply a common ratio

find r: r=a(sub)n+1/an

Alternating Sequences

goes from positive term to negative, pos, neg, pos, neg, …

Recursive Formula

formula: an=asub(n-1) +d

Explicit Formula for all Geometric

an=a1(r^n-1)

Recursive Formula for Geometric

an=a(sub)n-1 * r

Convergent Sequence

Given a sequence {an} if the terms an becomes artbitrarily close to a finite number L, as n becomes sufficiently lathe, we say {an} is a convergent sequence and L is the lmit of the sequence, In this case we write the lim as n goes to inifnity of an = L.

if the sequence is not convergent we say it is divergent

L’Hopital Rule

Used to evaluate limit of a function is indeterminate (0/0 or oo/oo or any combination), then you can use this rule to find the derivative of the numerator by itself, and the derivative of the denominator by itself. do not use quotient rule!!!

Finding horizontal asymptope of px^a/qx^b

if a>b, no HA

if b>a, HA @ 0

if a=b, HA @ p/q

Derivative of exponential function

Squeeze Theorem for Sequences

put sequence b in between sequence a and c to determine where sequence b converges or diverges

Bounded Above

if it is bounded above by a M, then M = y if your ceiling, and M is your horizontal aymptope

the shape of the sequence dots can only be horizontal or concave down

Bounded Below

if it is bounded below by a M, then M = y if your floor, and M is your horizontal aymptope

the shape of the sequence dots can only be horizontal or concave up

Bounded Sequence

the sequence has both a ceiling and a floor

boundary for {1/n}?

it is bounded above because 1/1 is 1 which 1 ≤ 1. we are looking at sequences

Relationship between boundaries and sequences

if a sequence converges, then it is bounded

if a sequence is bounded, it doesnt have to converge

Monotonic

an ≤ an+1

meaning that a number in a sequence has to be greater than or equal to the previous term

ex: {1, 2, 3, 4, 5, …} monotonic
ex: {1, 1, 2, 3, 4, …} monotonic
ex: {1, 0, -1, -2, …} monotonic
ex: {1, 2, 3, 2, 1, ….} not monotonic

Factorial

factorial is the number times all the following terms

ex: 5! = 5×4×3×2×1

ex: (2n+3)! = 2n!(2n+1)(2n+2)(2n+3)

n! = n (n-1) (n-2) (n-3) … * 1

Partial Sums

The sum of some of the terms in your sequence, not all of the terms

Infinite Series

A series that has an infinite number of terms

Finite Series

The sum of a finite number of terms

x^0

any number to a power of 0 is 0

Convergent Series

A series that adds up to a finite number or a sum that equals to a finite number

Divergent Series

The sum goes to infinity

1/n(n+1)

goes to 1

Geometric Series

goes to 2

if |r| < 1 then the series converges
if |r| ≥ 1 then the series diverges

Harmonic Series

diverges

Partial Sums Formula

works only on geometric series

Telescoping Series

first term - last term OR first half term - last half term

ex: {1 - 2} + {2 - 3} + … + {9 - 10}
1-10 = 9

Divergence Test (or nth term test)

If we have a series that converges then an must = 0 or a(infinity) = 0. it doesnt tell you if the series converges

Integral Test

P-Series

1. if it looks like a p-series
2. what is p?

Estimating the Remainder

1. make sure it satisfies the conditions

2. find the integral of f(x) from N to oo to find RN

Comparison Test

Compare one function we don’t know how to do with another function we know how to do

for any series an, we want to find a bn, such that bn is always less than an and greater than 0

0≤an≤bn OR an≥bn≥0

Limit Comparison Test

take an and divide it by bn, then apply the limit, if it

≠ 0 then both converge or diverge

= 0 then bn converges, then an converges
then bn diverges, then an is inconclusive

= ∞ then bn diverges, then an diverges
then bn converges, then an is inconclusive

Alternating Series

the terms alternate signs, from negative to positive and vice versa
(-1)^n/n to start with negative
(-1)^n+1/n to start with positive

three conditions
i. alternating series
ii. series is non-increasing
iii. lim of series bn = 0 (the last term is 0)

Sequence of odd numbers

ex: 1, 3, 5, 7, 9, = 2n+1

ex: 2, 4, 6, 8, 10 = 2n

Find Remainder in Alt. Series

do |RN| ≤ b(sub)N+1

plug in bN and plug in N

Absolute convergence

exhibits absolute convergence when if the series converges with |an|

Conditional convergence

exhibits conditional convergence if the series diverges with |an|

Alternating harmonic series

it is conditionally convergent

Ratio Test

take the series and do n+1 to every n, then divide the series by the original series

Root Test

Definition of e

lim as n goes to oo of (1 + 1/n)^n

Choosing a Convergence Test

Series Tests

a telescoping series is one in which most of the terms cancel in each of the partial sums

Power Series

Used to represent functions and define new functions. A function with a coefficient times a variable, where the variable has an exponent.

Can be thought of as an infinite polynomial (therefore always an infinite series) and also geometric series where r = x

∑n=0 to ∞ of c(sub)nx^n = c0X^0 + c1X^1 + c2X^2 + c3X^3 + …
∑n=0 to ∞ of c(sub)n (x-a)^n = c0(x-a)^0 + c1(x-a)^1 + c2(x-a)² + c3(x-a)³ + …

|x| ∠ 1 it converges
|x| ≥ 1 it diverges

Center location is the center of the power series graph
x-a; a is the center

Convergence of Power Series

Converges at x=a, diverges when x doesnt equal a
Converges for all real numbers it converges
Converges between a-R to a+R, outside of that it diverges and have to test the endpoints using the ratio test

Find radius of convergence

How to find radius:

Find interval of convergence

How to find interval of convergence: test both endpoints using the ratio test, put a-R into power series and use ratio test to see if the series converges or diverges. Repeat the process for a+R

if it converges, then change parenthesis to brackets
if it diverges, then keep the parenthesis

Conditions of power series

i. radius: 0. interval: [0,0]
ii. radius: ∞. interval: (-∞,∞)
iii. radius = a. interval (a-R, a+R) or [a-R, a+R]

Representing a Function using Power Series

If the function f(x) = 1/1-x then we can use the summation of n=0 to oo of x^n

Indicate the radius or interval of convergence

Your function is exactly like the infinite sum of a geometric series where the r=x and a=1. I can use a geometric series to represent it, since it is the sum of a geometric series formula (S∞ = a/1-r) → 1/1-r

Representing a function that can’t use geometric or power series directly

f(x) and identify a g(x) such that g’(x) = f(x). Then the power series representing of this f(x) is the derivative of the power series that represents g(x)

EX: g(x)=1/(1-x)² we don’t know how to represent this, so we identify f(x)=1/1-x we find that f’(x)=g(x)
1/1-x the power series is ∑n=0 to ∞ of x^n. derivative of this power series is ∑n=0 to ∞ of nx^(n-1) OR ∑n=0 to ∞ of (n+1)x^n
could be done using integral — we have h(x) = ∫k(x)dx

i

imaginary numbers

√(-1) even root of negative # = i
odd root of positive # = a certain #

Combining Power Series

when you combine two series with different intervals, use the smaller interval

Taylor Series Formula

Taylor Polynomial

A finite Taylor Series

0th Taylor poly. or 0th partial sum = f(a)
3rd taylor poly. or 3rd partial sum = psub3(x) = 4 terms