Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

- StudyBlue
- Australia
- University of Technology, Sydney
- Mathematics
- Mathematics 33360
- Various
- Review of Functions

Jacqui D.

Domain: -∞, ∞

Range: -1,1

Period (as it is periodic) : 2π

Range: -1,1

Period (as it is periodic) : 2π

Sin(0) = ?

0

Advertisement

Domain: -∞, ∞

Range, -1, 1

Range, -1, 1

cos(0) = ?

1

2sin(x)

Domain: -∞, ∞

Range: -2, 2

Domain: -∞, ∞

Range: -2, 2

1/2 sin(x)

Domain: -∞, ∞

Range: - 1/2, 1/2

Period: 2π

Domain: -∞, ∞

Range: - 1/2, 1/2

Period: 2π

sin(2x)

Domain: -∞, ∞

Range: -1, 1

Period: π

Domain: -∞, ∞

Range: -1, 1

Period: π

sin ( x/2 )

Domain: -∞, ∞

Range: -1, 1

Period: 4π

Domain: -∞, ∞

Range: -1, 1

Period: 4π

tan(x)

Domain: all real no.s except pi/2 + k pi

Range: all real no.s

Period: π

Domain: all real no.s except pi/2 + k pi

Range: all real no.s

Period: π

Trig. identities equivalent to sinθ

cos(π/2 - θ)

1/cscθ

1/cscθ

Trig. identities equivalent to cosθ

sin(π/2 - θ)

1/secθ

1/secθ

Advertisement

Trig. identities equivalent to tanθ

sinθ/cosθ

cot(π/2 - θ)

1 / cotθ

cot(π/2 - θ)

1 / cotθ

Trig. identities equivalent to cosecant or cscθ

sec(π/2 - θ)

1 / sinθ

1 / sinθ

Trig. identities equivalent to secant or secθ

csc(π/2 - θ)

1/cosθ

1/cosθ

Trig. identities equivalent to cotangent or cotθ

cosθ/sinθ

tan(π/2 - θ)

1/tanθ

tan(π/2 - θ)

1/tanθ

π/6 radians in degrees

30

π/4 radians in degrees

45

π/3 radians in degrees

60

π/2 radians in degrees

90

cos(0)

1

sin(0)

0

cos(π/6)

√3 / 2

sin(π/6)

1/2

cos(π/4)

1/√2

sin(π/4)

1/√2

cos(π/3)

1/2

sin(π/3)

√3/2

cos(π/2)

0

sin(π/2)

1

d/dx sinx

cosx

d/dx cosx

-sinx

sin^{2}θ + cos^{2}θ = 1

Rearrange this trig. identity into solutions for 1. (sin^{2}θ), 2.(cos^{2}θ) and 3.(sin^{2}θ - cos^{2}θ)

Rearrange this trig. identity into solutions for 1. (sin

1. (1-cos2θ)/2

2. (1+cos2θ)/2

3. -cos2θ

2. (1+cos2θ)/2

3. -cos2θ

sinx

cosx

sin^{-1} or arcsin

Domain: (-1,1) Range: (-π/2, π/2)

Derivative: 1/(√1-x^{2})

Domain: (-1,1) Range: (-π/2, π/2)

Derivative: 1/(√1-x

cos^{-1} or arccos

Domain: (-1,1) Range:(0,π)

Derivative: -1 / √(1-x^{2})

Domain: (-1,1) Range:(0,π)

Derivative: -1 / √(1-x

tan^{-1} or arctan

Domain: (-∞,∞) Range: (-π/2, π/2)

Derivative: 1 / (1+x^{2})

Domain: (-∞,∞) Range: (-π/2, π/2)

Derivative: 1 / (1+x

sinc x =

1 for x = 0

(sinx)/x for x ≠ 0

1 for x = 0

(sinx)/x for x ≠ 0

exponential or e^{x}

Domain: (-∞, ∞) Range: (0, ∞)

Derivative: e^{x}

Domain: (-∞, ∞) Range: (0, ∞)

Derivative: e

e^{0}=?

1

e^{1}=?

e ≅2.718

e^{x}e^{y}=?

e^{x+y}

(e^{x})^{y}=?

e^{xy}

e^{-x}=?

1/ e^{x}

lnx, Inverse of e^{x}

[Note: ln(e^{x}) = x and e^{lnx} = x]

[Note: ln(e

Gaussian

[Note: b is often replaced by μ and c by σ]

[Note: b is often replaced by μ and c by σ]

The exponential function, e^{x}

Gaussian

Norm. form:

g(x) = 1/[σ√(2π)] e^{- [(x-μ)^2 / 2σ^2]}

FWHM: 2√[2ln(2)]σ ≅ 2.35σ

Norm. form:

g(x) = 1/[σ√(2π)] e

FWHM: 2√[2ln(2)]σ ≅ 2.35σ

How do you form a gaussian function?

Gaussian functions arise by applying the exponential function to a general quadratic function. The Gaussian functions are thus those functions whose logarithm is a quadratic function.

What are gaussian functions useful for (applications)?

They are used in statistics (normal distribution) and physics (diffusion processes, optics of a laser beam).

Bessel function

Note: ∝ can be real or complex

Note: ∝ can be real or complex

Note: These functions are useful to describe 2D problems (e.g.: light scattering by rods)

Bessel function of second kind: Y_{a}(x) singular at the origin (x=0)

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly! I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2015 StudyBlue Inc. All rights reserved.

© 2015 StudyBlue Inc. All rights reserved.