Astronomy: Distance and Midpoint Between Two Celestial Objects

Astronomy: Distance and Midpoint Between Two Celestial Objects


Stars and celestial objects have the equatorial coordinates ( α, δ, R ) where:

α = right ascension in hours, minutes, and seconds
δ = declination in degrees, minutes, and seconds
R = distance to the celestial about (in light years, astronomical units, etc)

Convert α and δ to decimal degrees:

α = (hours + minutes/60 + seconds/3600) * 15
δ = sign(δ) * ( |degrees| + |minutes/60| + |seconds/3600| )

However, calculating distance between two points in space requires that we have cartesian coordinates (x, y, z). The conversion formulas to go from equatorial to cartesian coordinates are:

x = R * cos δ * sin α
y = R * cos δ * cos α
z = R * sin δ

For completeness, here are the conversion formulas to from cartesian to equatorial coordinates:

R = √(x^2 + y^2 + z^2)
δ = asin(z/R)
α = atan(y/x) (pay attention to the quadrant!). Alternatively: α = arg(x + i*y)

The distance formula between two points in space is:

D = √( (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 )

Source: http://en.m.wikipedia.org/wiki/Equatorial_coordinate_system

Example

Find the distance between Sagittarius A* (center of the Milky Way Galaxy) and the Andromeda Galaxy (M31).

Equatorial Coordinates:

From Wolfram Alpha: ( http://m.wolframalpha.com )

Sagittarius A*:
R = 24,824 light years
α = 17h45m40s = 266.42°
δ = -29°0'28" = -29.008°

Andromeda Galaxy:
R = 2,571,000 light years
α = 42m40s = 10.68°
δ = 41°16'8" = 41.269°

Converting to the cartesian system:

Sagittarius A*:
x = -21,667.514038
y = -1,355.611275
z = -12,037.945401

Andromeda Galaxy:
x = 358,122.65578
y = 1,898,943.67736
z = 1,695,818.99789

Calculation:

D = √( (-21,667.514038 - 358,122.65578)^2 + (-1,355.611275 - 1,898,943.67736)^2
+ (-12,037.945401 - 1,695,818.99789)^2 ) ≈ 2,583,051.16059

The distance between Sagittarius A* and the Andromeda Galaxy is about 2.583 million light years.

Happy Thanksgiving, or Day of Thanks, wherever you are. Please be safe and use good judgement. We need all the positive we can get on Earth, and in the cosmos. Cheers!

Eddie

This blog is property of Edward Shore. 2014

Complex Analysis: Line Integral

Contour Integrals

Let the complex function f(z) be defined as f(z) = u(x,y) + i*v(x,y)

Using the definition of the integral:

∫ f(z) dz = lim n → ∞ [ Σ f(z_k) * Δz_k from k = 1 to n ]
= lim n → ∞ [ Σ ( u(x_k, y_k) + i*v(x_k, y_k) ) * ( Δx_k + i*Δy_k ) from k = 1 to n ]
= lim n → ∞ Σ [ (u(x_k, y_k) * Δx_k - v(x_k, y_k) * Δy_k) + i*(v(x_k, y_k) * Δx_k
+ u(x_k, y_k) * Δy_k) ]
= ( ∫ u(x,y) dx - ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )

In summary:

---

∫ f(z) dz = ( ∫ u(x,y) dx - ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )

---

What is needed:

1. A contour curve f = y(x). This is where you get your end points.

2. We need f(z) to integrate. Separate f(z) into its parametric parts.

Let's look at a couple examples.

Examples

1. Contour: y = x + 1 from (3,4) to (4,5). Integrate ∫ z^2 dz.

Since y = x + 1, x = y - 1

z^2 = (x + i*y)^2 = (x^2 - y^2) + i*(2*x*y)
u = x^2 - y^2
v = 2*x*y

u = x^2 - y^2
u = x^2 - (x + 1)^2
u = -2*x - 1
∫ u dx from 3 to 4 = -8

v = 2 * x * y
v = 2 * (y - 1) * y
∫ v dy from 4 to 5 = 95/3

u = (y - 1)^2 - y^2
u = -2 * y + 1
∫ u dy from 4 to 5 = -8

v = 2 * x * y
v = 2 * x * (x + 1)
∫ v dx from 3 to 4 = 95/3

∫ z^2 dz from (3 + 4*i) to (4 + 5*i) = (-8 - 95/3) + i*(-8 + 95/3) ≈ -39.666667 + 23.666667*i

2. Contour: y = x^2 - 1 from (0, -1) to (2, 1). Integrate ∫ z^2 + 1 dz.

y = x^2 - 1, x = √(y + 1)

z^2 + 1 = (x + i*y)^2 + 1 = x^2 + 2*i*x*y - y^2 + 1
u = x^2 - y^2 + 1
v = 2 * x * y

u = x^2 - y^2 + 1
u = x^2 - (x^2 - 1)^2 + 1
∫ u dx from 0 to 2 = 8/5

v = 2 * x * y
v = 2 * √(y + 1) * y
∫ v dy from -1 to 1 = 8/15 * √2

u = x^2 - y^2 + 1
u = y + 1 - y^2 + 1
u = -y^2 + y + 2
∫ u dy from -1 to 1 = 10/3

v = 2 * x * y
v = 2 * x * (x^2 - 1)
∫ v dx from 0 to 2 = 4

∫ z^2 + 1 dz from -i to 2+i = (8/5 - 8/15 * √2) + i*(4 + 10/3) ≈ 0.845753 + 7.333333*i

This is the basics of line integrals.

Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. Addison-Wesley Publishing Company. 1994.

Take care, Eddie

This blog is property of Edward Shore. 2014

Complex Analysis: Cauchy-Reimann Equations

Continuing my blogging adventures at the Last Drop Cafe in Claremont, CA:

Cauchy-Reimann Equations

Let the complex-valued function f(z), where z = x+i*y, be defined in the parametric form:

f(z) = u(x,y) + i*v(x,y) where u(x,y) = Re(f(z)) and v(x,y) = Im(f(z)).

Then by the definition of the derivative:

f(z) = lim x0 → x [(f(z) + x0) - f(z))/(x - x0)]
= lim x0 → x [(u(x +x0, y) - u(x, y) + i*v(x + x0, y) - i*v(x, y))/(x - x0)]
= du/dx + i*dv/dx (I)

Also:

f(z) = lim y0 → y [(f(z + y0) - f(z))/(i*y - i*y0)]
= lim y0 → y [(u(x, y + y0) - u(x, y) + i*v(x, y + y0) - i*v(x, y))/(i*(y - y0))]
= -i*(lim y0 → y [(u(x, y + y0) - u(x, y) + i*v(x, y + y0) - i*v(x, y))/(y - y0)])
= -i*du/dy + dv/dy (II)

Taking the real and imaginary parts of (I) and (II):

du/dx = dv/dy and -du/dy = dv/dx

The Cauchy-Riemann equations can be used to determine whether f(z) is differentiable and if so, where.

Two Quick Examples

1. z^2 = (x^2 - y^2) + 2*i*x*y

u=x^2 - y^2
v=2*i*x*y

du/dx = 2*x, dv/dy = 2*x
-du/dy = 2*y, dv/dx = 2*y (note the negative sign in front of du/dy!)

Since the Cauchy-Riemann equations hold, and without restriction, then z^2 is differentiable for all z. And:

df/dz = 2*x + 2*i*y = 2*(x+i*y) = 2*z

2. e^z = e^(x+i*y) = e^x*cos y + i*e^x*sin y

u=e^x*cos y
v=e^x*sin y

du/dx = e^x*cos y, dv/dy = e^x*cos y
-du/dy = e^x*sin y, dv/dx = e^x*sin y

Since the Cauchy-Riemann equations hold, and without restriction, then e^z is differentiable for all z. And:

df/dx = e^x*cos y + i*e^x*sin y = e^z

I used (I), but using (II) will garner the same result.

That is the Cauchy-Riemann equations in a nutshell!

Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. Addison-Wesley Publishing Company. 1994.

As always, have a great day! Take care. Eddie

This blog is property of Edward Shore. 2014

Complex Analysis: The Conjugate, the Modulus, and its Properties

Blogging today from Last Drop Cafè in Claremont, CA. I think I found a new favorite drink: mint mocha made with soy milk.


Here are some basics of the conjugate and modulus of complex numbers.

Let z = x+i*y, with i=√-1

Conjugate

Usually labeled "z-bar" (z with a line over it), the conjugate is also labeled conj(z) and z*.

conj(z) = x - i*y

Modulus or Absolute Value

Not surprisingly, the modulus, also called the absolute value of the complex number z is defined as:

|z| = √(x^2 + y^2)

Properties

Let's explore some properties of the conjugate and modulus.

z + conj(z) = (x + i*y) + (x - i*y) = 2*x

z - conj(z) = (x + i*y) - (x - i*y) = 2*i*y

(conj(z))^2 = (x - i*y)^2 = x^2 - 2*i*x*y + (i*y)^2 = x^2 - y^2 - 2*i*x*y

conj(z)^2 + z^2 = (x - i*y)^2 + (x + i*y)^2 = x^2 - 2*i*x*y - y^2 + x^2 + 2*i*x*y - y^2
= 2*(x^2 - y^2)

conj(z) * z = (x - i*y) * (x + i*y) = x^2 + i*x*y - i*x*y - i^2*y^2 = x^2 + y^2 = |z|^2

which easily leads to: |z| = √(z * conj(z)) and

|z1 * z2| = √(z1 * conj(z1) * z2 * conj(z2)) = √(z1 * conj(z1)) * √(z2 * conj(z2)) = |z1| * |z2|

and:

|z1/z2| = √((z1 * conj(z1))/(z2 * conj(z2))) = √((z1 * conj(z1))/√((z2 * conj(z2)) = |z1|/|z2|


Source: Wuncsh, David A. Complex Numbers with Applications. 2nd Edition. Addison-Wesley Publishing Company. Reading, MA. 1994

Casio fx-9750gII, fx-9860gII, Prizm: Prime and Complex Number Prime (Gaussian Prime)

Let /rtri be the right triangle symbol (display symbol)

Real Number Primes

Program ISPRIME:

“REAL PRIME”

“EWS 2014-11-12”
“N”?->N
Int N->N
For 2->K To S
If Frac (N÷K)=0
Then
“NOT PRIME” /rtri
Stop
IfEnd
Next
“PRIME”

 Examples:

19 is PRIME
36 is NOT PRIME

Complex Number Primes

(Gaussian Primes)

Source:  http://mathworld.wolfram.com/GaussianPrime.html

Program CISPRIME:

“COMPLEX PRIME”

“EWS 2014-11-12”
a+bi
“Z”?->Z
(Abs Z)^2->T
If ReP Z=0 or ImP Z=0
Then
Goto 1
IfEnd
For 2->K To Abs Z
If Frac(T÷K)=0
Then Goto 2
IfEnd
Next
Goto 3
Lbl 1
If MOD(Abs Z, 4)≠3
Then Goto 2
IfEnd
Next
Goto 3
Lbl 2
“NOT PRIME” /rtri
Stop
Lbl 3
“PRIME”

Examples:

2+2i is not (complex) prime
7 is (complex) prime
-5+4i is (complex) prime

 

 

This blog is property of Edward Shore.  2014

 

Blog Entry #404

 

 

Darts: The Probability of Getting the Best Score

Darts: The Probability of Getting the Best Score

In Darts, players throw darts at circular board.  The closer you are to the center of the circle, the more points you score.  Hitting the center of the circle (or close to it) scores you the bulls-eye.    

Today we will develop formulas that determine the chance of landing a dart in a specific scoring zone.  For this, we will assume that all the darts thrown will land in a specific area.  The dart that hits the cat who are unfortunate to be in its path don’t count.

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Area of a Circular Ring

We will be using the area of a circular ring heavily in this blog.  The general formula for it is:

 A =  π * ((outer radius)^2 – (inner radius)^2)

---

 

Simplified Darts Board




 





This board has a circular area on a square board.  The square board has side length x.  There are three circular scoring areas:

(I)                 Bulls-eye circle with radius R1.  Obviously, this area has the most points.

(II)               Green scoring area, a circular ring with outer radius R2 and inner radius R1. This area is second best in points to the bulls-eye.

(III)             Yellow scoring area, a circular ring with outer radius R3 and inner radius R2. 

A dart that lands outside any of the scoring areas scores no points. 

Assuming all but a negligible of parts hits the square board, the chance that a dart hits the bulls-eye is:

P1 = (area of the bulls eye)/(area of the board) = π*R1^2/x^2

The probability that a dart hits the green scoring area:

P2 = (area of the green scoring area)/(area of the board) = π*(R2^2 – R1^2)/x^2

The probability that a dart hits the yellow scoring area:

P3 = (area of the yellow scoring area)/(area of the board) = π*(R3^2 – R2^2)/x^2

Finally, the probability of scoring anything at all on this board is:

P_any = P1 + P2 + P3 = π*P3^2/x^2

This assumes that the dart board is flat and each point of the dart board has an equal random chance.  Note that these probability formulas do not take into account the skill of the dart thrower. 

Standard Darts Board

 

 

The standard darts board is a circular board with radius R. Going from the inside out, we have:

 

(I)                 Inner Bulls-eye.  In the standard game, the inner bulls-eye is 50 points.  It is a circular area, with radius R1.  

(II)               Outer Bulls-eye – worth 25 points.  It is a circular ring, with outer radius R2 and inner radius R1. 

(III)             Inner Single Ring.  From here on out, the board is divided into 20 zones, with each zone assigned a different amount of points from 1 to 20 (not shown on the diagram).  Here I am only concerned with the circular ring in general.  This ring has the outer radius of R3 and inner radius of R2.

(IV)             Triple Ring.  Each of the zones in the triple ring are worth triple the points of designated zone.   Let’s assign the outer radius of the triple ring R4 and inner radius, R3.

(V)               Outer Single Ring.  With outer radius R5, and inner radius R6.

(VI)             Double Ring.  Each of the zones in the double ring are worth double the points.   This ring has the outer radius R6 and inner radius R5.

 

Here, I am concerned about hitting the general scoring areas. 

Probability of hitting a single scoring area:

P1 = (area of inner single ring + area of outer single ring)/(area of the board)

= (π * (R3^2 – R2^2) + π * (R5^2 – R6^2))/(π *R^2)
= (π * (R3^2 – R2^2 + R5^2 – R6^2)/(π * R^2)
= (R3^2 – R2^2 + R5^2 – R6^2)/R^2

Probability of hitting a double ring:

P2 = (area of the double ring)/(area of the board)

= (π * (R6^2 – R5^2))/(π * R^2)
= (R6^2 – R5^2)/R^2

Probability of hitting a triple ring:

P3 = (area of the triple ring)/(area of the board)

= (π * (R4^2 – R3^2))/(π * R^2)
= (R4^2 – R3^2)/R^2

Probability of hitting the bulls-eye:

PB = (area of the inner bulls-eye + outer bulls-eye)/(area of the board)

= (π * R1^2 + π * (R2^2 – R1^2))/(π * R^2)
= (π * R2^2)/(π * R^2)
= R2^2/R^2

Scoring 60

The highest a single dart can score is 60, hitting the 20 zone in the triple ring.  Note the triple ring is divided into 20 zones of equal area.  The required probability of scoring 60 points is:

P60 = (R4^2 – R3^2)/20 * 1/R^2 = (R4^2 – R3^2)/(20*R^2)

Knowing the area of the circular ring is key to calculating probabilities of hitting areas of the dart board.

Source:  http://en.wikipedia.org/wiki/Darts

Eddie

This entry is property of Edward Shore. 2014.

 

Two New Videos: Derivatives on the HP Prime, Derivative and Integral of abs(x)

Derivatives on the HP Prime Calculator

Video:  http://youtu.be/LXwLiPlKO2M

Notes:

Derivative Template:  Template Key, template is on the top row, 4th column

Numeric Format:  (d* function)/(d varaible = point), press Enter
In RPN Mode: You also need to press Shift+Comma (Eval)
Symbolic Format (CAS Mode):  (d function)/(d variable)
*d represents the derivative symbol

diff command:
diff(function, variable)

Derivative and Integral of abs(x)

Video:  http://youtu.be/8PikrlA5UlQ

A short derivation of the derivative and integral of absolute value function (abs(x)), and how the signum (sign(x)) function plays a part its calculation. 

Graphs made with the online Desmos calculator.  Please visit and explore at www.desmos.com
(This is NOT a paid endorsement)

Eddie


Blog Entry #403

This blog is property of Edward Shore.  2014

My Morning at Honnold/Mudd Library: 2014-11-08 (Game Shows, Bravias Lattice, Prime Numbers)

My Morning at Honnold/Mudd Library

Honnold/Mudd Library - Claremont, CA

What I love about visiting the mathematics and science sections in a university library is that there is a feast of information.  Every time, there is something new to learn.   Here are some notes from today’s visit:

Game Shows and Winning

Source:  John Haigh.  “Taking Chances: Winning with Probability”  2^nd Edition Oxford Press: 2003

Haigh introduces two central ideas of probability:  exclusive events and independence.  Is it impossible for two events to occur simultaneously?  Does one event have any effect on another?  Let’s explore some popular concepts of probability found in game shows.

“Find the Lady” – also known as the “Monty Hall Problem”.  You have three boxes, two are worthless and one has the prize.  After the contestant selects the box, the host shows one of the boxes that are worthless.  Statistically, if the contestant switches their initial choice, the contestant doubles the chance of winning.  That is, since the contestant initially is incorrect 2/3 of the time, and the host shows one of the incorrect boxes; the contestant who switches has a 2/3 chance of switching from a losing box to the winning box.

“Showcase Showdown” (The Price is Right) – spinning last may not be the most advantageous place to be.  It depends on the scores of the first two spinners, and the chance of getting the best score. 

“Who Wants to be a Millionaire” – In the classic structure of the game, before the time clock and the shuffle format, contestants were offered three lifelines:  Ask the Audience, 50:50, and Phone a Friend.  Haigh suggests using the lifelines in this order: 50:50, Ask the Audience, and Phone a Friend.  The risk between higher level questions increase as the game goes on.  

“Blockbusters”

A two-person team (white) goes against an individual player (blue in the UK, red in the US).  For each question the team has 2/3 chance of getting the question correctly, but will have to answer a minimum of five question correctly, while the individual can win the game as a few as four correct answers.

My calculation:  assuming that there are no blocks, the probability that the team wins with five consecutive answers is (2/3)^5 = .13169 versus (1/3)^4 = .01235. 

Ideal Crystals and Bravais Lattices

Source:  M.G. Cottam and D.R. Tilley  “Introduction to Surface and Superlattice Excitations”  Cambridge University Press:  Cambridge.  1989.

Ideal crystals are described as a basis of atoms (or ions) located on each point of a lattice, which is defined as a regular periodic array in space.  In an ideal crystal, all the lattice points are equivalent.

Each lattice is defined by the non-coplanar vector a1, a2, and a3.  That is, a1, a2, and a3 do not lie on one specific plane.  The space lattice is formed through a collection of vectors, symbolized as R, as:

R = n1*a1 + n2*a2 + n3*a3, where n1, n2, and n3 are integers. 

Since the ideal crystal has translational symmetry, the transition f(r+R)=f(r) is satisfied for all points in space, which such a point is represented by r. 

The reciprocal lattice vector of R, symbolized by Q, takes the form:

Q = v1*b1 + v2*b2 + v3*b3, where v1, v2, and v3 are integers, and

b1 = ((2*π*a2) x a3) / dot(a1, a2 x a3)

b2 = ((2*π*a3) x a1) / dot(a1, a2 x a3)

b3 = ((2*π*a1) x a2) / dot(a1, a2 x a3)

Where x represents the vector cross product and dot(v,u) represents the dot product.

Two-Dimensional Lattices

Let a1 and a2 be two translation vectors and the set R be defined as:

R = n1*a1 + n2*a2, where n1 and n2 are integers

The ends points of R from what is called the Bravais lattice.  There are five Bravais lattices which take the shape of one of these five forms:

1.    Square, |a1| = |a2|, θ = 90°

2.    Primitive Rectangle (general rectangle), |a1| ≠ |a2|, θ = 90°

3.    Centered Rectangle (rhombic), |a1| ≠ |a2|, θ = 90ׄ°

4.    Hexagonal, |a1| = |a2|, θ = 60° or 120°

5.    Oblique, all other cases

Cottam and Tilley illustrates the concept of bulk and surface excitation by using how Rayleigh’s descriptions of waves in a semi-infinite isotropic elastic medium.  In searching for plane-wave solution, in which the wave propagates parallel to the surface with vector q with velocity v and frequency w:

f’’(z) – (q^2 – w^2/v^2)*f(z) = 0

If q^2 < w^2/v^2, we have a bulk wave where

f(z) = B1*e^(i*√(w^2/v^2-q^2)*z) + B2*e^(-i*√(w^2/v^2-q^2)*z) where B1 and B2 are constants.

If q^2 > w^2/v^2, we have a decaying surface wave where

f(z) = B3*e^(√(q^2-w^2/v^2)*z)

Prime Number Goodies

Source:  Richard Carndall and Carl Pomeance.  “Prime Numbers: A Computational Perspective.  Second Edition” Springer.  2005

Twin Primes:  

Two primes that differ by 2.  Examples include 3 and 5, 29 and 31, and 311 and 313.  Is there an upper bound on twin primes, or is there an infinite set of pairs? 

V. Brun showed the upper bound for the number of twin prime sets under the integer x is:

‘π(x) = O(x*(ln ln x/ln x)^2) -> x*(ln x)^-2

Brun also showed that the sum of the reciprocal of the twin primes is finite.  That sum is denoted as the Brun constant which is:

B = (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + …

The Brun constant is at least 1.71077 (using twin primes from 1 to 1,000,000).  As imagined, this series converges slowly.

Formulas to Produce Primes:

Euler used the polynomial x^2 + x + 41 to produce primes, which worked for all integers 0 to 39.  After x = 39, the polynomial continues produce primes at a remarkable probability. 

Mersenne Primes:

Mersenne primes, in the form of 2^q – 1 where q is a prime greater than 2.  It has been proven that if 2^q – 1 is prime, then so is q.  This form does not always produce primes (see q = 11 or q = 23).  In fact, the distribution of Mersenne primes gets scarce quickly as q increases.

Perfect numbers, integers in which the sum of its divisors except itself equals that integer (example: 6 = 1 + 2 + 3), can be characterized by Euler and Euclid:

An even number is perfect if and only if n = 2^(q-1)*(2^q-1) where 2^q-1 is prime.

Quadratic Residues:

The integer a is said to be a quadratic residue of coprime integer m if the congruence

x^2 ≡ a mod m

can be solved such as x is an integer and gcd(a, m) = 1. 

For example, 19 is a quadratic residue mod 5 since 19 mod 5 = 4 = 2^2 and gcd(19,5) = 1.

The next two basic definitions play a role in solving polynomials and equations involving prime numbers.

Legendre Symbol defined as (a/p) [normally written vertically]:

Let p be an odd prime.  Then:

(a/p) = legendresymbol(a,p) =

0 if a ≡ 0 mod p

1 if a is a quadratic residue of p

-1 if a is not a quadratic residue of p

Example:  p = 19.  The Legendre symbols of (4/19), (5/19), and (19/19) are 1, -1, and 0, respectively. (corrected on 2014-11-10 - with apologizes)

Jacobi Symbol defined as (a/m) [normally written vertically]:

Let m be any odd integer and a be any integer.  Then:

(a/m) = jacobisymbol(a, m) =  product( legendresymbol( a, prime factor of m )).

Example:  m = 27 = 3 * 9, a = 55

Jacobisymbol(55, 27) =

Legendresymbol(55, 3) * legendresymbol(55, 9) = 1 * 1 = 1

Is it shown that jacobisymbol(a, m) = 0 if and only if gcd(a, m) > 1.

Enjoy!  

Eddie

Blog Entry # 402

This blog is property of Edward Shore.  2014

Derivative of abs(x) and Integrals of abs(x), abs(e^x), abs(e^(a*x) + e^(a*y))

Derivative of abs(x) and Integrals of abs(x), abs(e^x), abs(e^(a*x) + e^(a*y))

For this blog entry, assume that are functions are in terms of x and a, x, y represent real numbers.

Definition of abs(x) (also symbolized as |x|)

Piecewise definition:

abs(x) = x if x > 0
abs(x) = 0 if x = 0
abs(x) = -x if x < 0

Also:

abs(x) = x * sign(x)

Where the sign(x), sometimes labeled sgn(x), is the sign or signum function. It is defined as:

sign(x) = 1 for x > 0
sign(x) = 0 for x = 0
sign(x) = -1 for x < 0

d/dx sign(x)

We can clearly demonstrate that d/dx sign(x) = 0 since:

d/dx sign(x) = 0 for x > 0
d/dx sign(x) = 0 for x = 0
d/dx sign(x) = 0 for x < 0

d/dx abs(x)

Using the chain rule:
d/dx abs(x)
= d/dx (x * sign(x))
= d/dx (x) * sign(x) + x * d/dx (sign(x))
= sign(x)

This can also be done with the piecewise representation:

d/dx abs(x) = 1 if x > 0
d/dx abs(x) = 0 if x = 0
d/dx abs(x) = -1 if x < 0

∫ abs(x) dx

∫ abs(x) dx
= ∫ x * sign(x) dx

Using integration by parts:
where u = sign(x), dv = x dx
Then: du = 0 dx, v = x^2/2

∫ abs(x) dx
= ∫ x * sign(x) dx
= x^2/2 * sign(x) - ∫ 0 dx
= x^2/2 * sign(x) + C

C is the arbitrary integration constant

With the piecewise representation:

∫ abs(x) dx = x^2/2 if x > 0
∫ abs(x) dx = 0 if x = 0
∫ abs(x) dx = -x^2/2 if x <0

Note that this is x^2/2 * sign(x).

∫ abs(e^(a*x)) dx

∫ abs(e^(a*x)) dx
= ∫ e^(a*x) * sign(e^(a*x)) dx
= 1/a * e^(a*x) * sign(e^(a*x)) - ∫ 0 dx
= 1/a * e^(a*x) * sign(e^(a*x)) + C

∫ abs(e^(a*x) + e^(a*y)) dx

∫ abs(e^(a*x) + e^(a*y)) dx
= ∫ (e^(a*x) + e^(a*y)) * sign(e^(a*x) + e^(a*y)) dx
= (e^(a*x)/a + x*e^(a*y)) * sign(e^(a*x) + e^(a*y)) - ∫ 0 dx
= (e^(a*x) + a*x*e^(a*y))/a * sign(e^(a*x) + e^(a*y)) + C

Eddie

This blog is property of Edward Shore. 2014

(Blog Entry # 401)