different S-shaped curves

different S-shaped curves

I’M ALIVE. HAVE AN ART DUMP.

From top to bottom:

Sech(z), real part purple, imaginary part yellow.

ArcTan(z^4)+ArcSech(z^2), real part purple, imaginary part green.

Csc(z^2)+ArcCot(z^2), real part teal, imaginary part pink. In memory of Dave Grimes.

ArcSech(z), real part blue, imaginary part pink.

ArcSinh (z^4)+ArcTan (z^2), real part blue, imaginary part yellow.

Csch(z^4)+ArcSech(z^2), real part purple, imaginary part teal.

The Inverse Trigonometric Functions

**Definition of Inverse and One-to-one Functions**

An inverse function is a defined function that is found using its original function. In order for a function’s inverse to be defined, the function itself must be injective, or one-to-one, which means it must have only one x value for every y value. If the function is non-injective, or not one-to-one, then the domain must be restricted before it can have its inverse defined.

When defined on a whole real line, the trigonometric functions are not one-to-one functions, making them not invertible. However, one can suitably restrict each trigonometric function’s domain so that they become a one-to-one function. The inverse trigonometric functions are the inverses of the trigonometric functions with restricted domains.

**Principal-value Sine Function**

The principal-value sine function, denoted as Sine or Sin, is the restriction of sine to the interval [-π/2,π/2].

Sine is a one-to-one function with an inverse function called Arcsine. Arcsine is also denoted as Arcsin or sin⁻¹. Sin is an increasing function.

domain of Arcsin = [-1,1] = range of Sin

range of Arcsin = [-π/2,π/2] = domain of Sin

The graph of y = Arcsinx is the mirror image of the graph of y = Sinx in the line y = x. Similarly, the graph of x = Siny is the mirror image of the graph of x = Arcsiny in the line y = x. Note that y = Arcsinx is equal to x = Siny, and x = Arcsiny is equal to y = Sinx.

**Derivative of Arcsin**

Implicit differentiation is used to calculate the derivative of Arcsin.

Since y = Arcsinx is equal to x = Siny:

Since cosy is positive on the unit circle between -π/2 and π/2, then the following is true:

Since x = siny, then:

Therefore, the following is the derivative of Arcsin, using above and the chain rule:

**Indefinite Integral of Arcsin**

The following is the derivative of Arcsin if x was multiplied by 1/|a| and 1/a (a > 0):

The first derivative produces the following indefinite integral:

**Principal-value Cosine Function**

The principal-value cosine function, denoted as Cosine or Cos, is the restriction of cosine to the interval [0,π].

Cosine is a one-to-one function with an inverse function called Arccosine. Arccosine is also denoted as Arccos or cos⁻¹. Cos is a decreasing function.

domain of Arccos = [-1,1] = range of Cos

range of Arccos = [0,π] = domain of Cos

The graph of y = Arccosx is the mirror image of the graph of y = Cosx in the line y = x. Similarly, the graph of x = Cosy is the mirror image of the graph of x = Arccosy in the line y = x. Note that y = Arccosx is equal to x = Cosy, and x = Arccosy is equal to y = Cosx.

**Derivative of Arccos**

Implicit differentiation is used to calculate the derivative of Arccos.

Since y = Arccosx is equal to x = Cosy:

Since siny is positive on the unit circle between 0 and π, then the following is true:

Since x = cosy, then:

Therefore, the following is the derivative of Arccos, using above and the chain rule:

**Indefinite Integral of Arccos**

The following is the derivative of Arccos if x was multiplied by 1/|a| and 1/a (a > 0):

The first derivative produces the following indefinite integral:

**Arcsin and Arccos Identities**

For -1 < x < 1:

Therefore, Arcsinx + Arccosx equals a constant on (-1,1).

Since Arcsin0 = 0 and Arccos0 = π/2, for any x in (-1,1):

Arcsinx + Arccosx = π/2

If -π/2 ≤ x ≤ π/2, then:

**Principal-value Tangent Function**

The principal-value tangent function, denoted as Tangent or Tan, is the restriction of tangent to the interval (-π/2,π/2).

Tangent is a one-to-one function with an inverse function called Arctangent. Arctangent is also denoted as Arctan or tan⁻¹. Tan is an increasing function.

domain of Arctan = (-∞,∞) = range of Tan

range of Arctan = (-π/2,π/2) = domain of Tan

The graph of y = Arctanx is the mirror image of the graph of y = Tanx in the line y = x. Similarly, the graph of x = Tany is the mirror image of the graph of x = Arctany in the line y = x. Note that y = Arctanx is equal to x = Tany, and x = Arctany is equal to y = Tanx.

**Derivative of Arctan**

Implicit differentiation is used to calculate the derivative of Arctan.

Since y = Arctanx is equal to x = Tany:

Since sec²y is positive, then the following is true:

Since x = tany, then:

Therefore, the following is the derivative of Arctan, using above and the chain rule:

**Indefinite Integral of Arctan**

The following is the derivative of Arctan if x was multiplied by 1/a:

This produces the following two indefinite integrals:

**Principal-value Secant Function**

The principal-value secant function, denoted as Secant or Sec, is the restriction of secant to 1/Cosx’s restriction, which is the same restriction as Cosx [0,π], however 1/Cosx is undefined at x = π/2, therefore Secant is the restriction of secant on [0,π/2) U (π/2,π].

Secant is a one-to-one function with an inverse function called Arcsecant. Arcsecant is also denoted as Arcsec or sec⁻¹.

domain of Arcsec = (-∞,-1] U [1,∞) = range of Sec

range of Arcsec = [0,π/2) U (π/2,π] = domain of Sec

The graph of y = Arcsecx is the mirror image of the graph of y = Secx in the line y = x. Similarly, the graph of x = Secy is the mirror image of the graph of x = Arcsecy in the line y = x. Note that y = Arcsecx is equal to x = Secy, and x = Arcsecy is equal to y = Secx.

**Derivative of Arcsec**

Arccos’s derivative is used to calculate the derivative of Arcsec.

**Indefinite Integral of Arcsec**

Coming soon…

**Principal-value Cosecant Function**

The principal-value cosecant function, denoted as Cosecant or Csc, is the restriction of cosecant to 1/Sinx’s restriction, which is the same restriction as Sinx [-π/2,π/2], however 1/Sinx is undefined at x = 0, therefore Cosecant is the restriction of cosecant on [-π/2,0) U (0,π/2].

Cosecant is a one-to-one function with the inverse function called Arccosecant. Arccosecant is also denoted as Arccsc or csc⁻¹.

domain of Arccsc = (-∞,-1] U [1,∞) = range of Csc

range of Arccsc = [-π/2,0) U (0,π/2] = domain of Csc

The graph of y = Arccscx is the mirror image of the graph of y = Cscx in the line y = x. Similarly, the graph of x = Cscy is the mirror image of the graph of x = Arccscy in the line y = x. Note that y = Arccscx is equal to x = Cscy, and x = Arccscy is equal to y = Cscx.

**Derivative of Arccsc**

Arcsin’s derivative is used to calculate the derivative of Arccsc.

**Indefinite Integral of Arccsc**

Coming soon…

**Principal-value Cotangent Function**

The principal-value cotangent function, denoted as Cotangent or Cot, is the restriction of cotangent to the same interval as Cosx, which is [0,π], however Cot is undefined when x = 0, π, therefore Cotangent is the restriction of cotangent on (0,π).

Cotangent is a one-to-one function with the inverse function called Arccotangent. Arccotangent is also denoted as Arccot or cot⁻¹.

domain of Arccot = (-∞,∞) = range of Cot

range of Arccot = (0,π) = domain of Cot

The graph of y = Arccotx is the mirror image of the graph of y = Cotx in the line y = x. Similarly, the graph of x = Coty is the mirror image of the graph of x = Arccoty in the line y = x. Note that y = Arccotx is equal to x = Coty, and x = Arccoty is equal to y = Cotx.

**Derivative of Arccot**

Arctan’s derivative is used to calculate the derivative of Arccot.

**Indefinite Integral of Arccot**

Coming soon…

**Evaluate the following expression:**

Since Arcsinx = y, Siny = x:

**Evaluate the following expression:**

**Arctan(-1)**

Since Arctanx = y, Tany = x:

Arctan(-1) = y <–> Tany = -1 = -π/4

Note Tan’s domain on the unit circle.

**Evaluate the following expression:**

Evaluating tan(2π/3):

Since Arctanx = y, Tany = x:

Note Tan’s domain on the unit circle.

**Evaluate the following expression:**

Let t = Arccosx, then if t = Arccosx, x = cost.

**Evaluate the following:**

**tanArctan200**

Let t = Arctanx, then if t = Arctanx, x = tant.

tant = x = 200

**Simplify the following expression:**

**cosArccosx**

Let t = Arccosx, then if t = Arccosx, x = cost.

cost = x = x

**Simplify the following expression:**

**sinArccosx**

Let t = Arccosx, then if t = Arccosx, x = cost.

**Simplify the following expression:**

**tanArccosx**

Let t = Arccosx, then if t = Arccosx, x = cost and sint = √(1 - cos²t) = √(1 - x²).

**Simplify the following expression:**

**Arcsinsinx**

The following is the graph of y = Arcsinsinx:

Notice that there are intervals of which Arcsinsinx is increasing and decreasing.

The following formula covers all intervals of x when Arcsinsinx is increasing:

The reason there is 2kπ is because it takes a period 2kπ to go from the next interval of x when Arcsinsinx is increasing.

The reason π/2 is subtracted from the left side and π/2 is added on the right side is because the starting interval of all intervals of x when Arcsinsinx is increasing is (-π/2,π/2). Note that it takes a period of π to go from one increasing interval to the next.

The following formula covers all intervals of x when Arcsinsinx is decreasing:

The reason there is 2kπ is because it takes a period 2kπ to go from the next interval of x when Arcsinsinx is decreasing.

The reason π/2 is added on the left side and 3π/2 is added on the right side is because the starting interval of all intervals of x when Arcsinsinx is decreasing is (π/2,3π/2), noting that π/2 is the ending point to the interval of when Arcsinsinx is increasing. Also note that it takes a period of π to go from one decreasing interval to the next.

Since Sinx’s domain is [-π/2,π/2]:

Then, Arcsinsinx = ArcsinsinX = Arcsinsin(x - 2kπ) = x - 2kπ

Then, Arcsinsinx = ArcsinsinX = Arcsinsin(2kπ + π - x) = 2kπ + π - x

Note that the laws of inequalities was used.

Therefore, there are two possible answers of simplifying Arcsinsinx:

**Simplifying the following expression:**

**Arccoscosx**

The following is the graph of y = Arccoscosx:

Notice that there are intervals of which Arccoscosx is increasing and decreasing.

The following formula covers all intervals of x when Arccoscosx is increasing:

The reason there is 2kπ is because it takes a period 2kπ to go from the next interval of x when Arccoscosx is increasing.

The reason nothing is subtracted from the left side and π is added on the right side is because the starting interval of all intervals of x when Arccoscosx is increasing is (0,π). Note that it takes a period of π to go from one increasing interval to the next.

The following formula covers all intervals of x when Arccoscosx is decreasing:

The reason there is 2kπ is because it takes a period 2kπ to go from the next interval of x when Arccoscosx is decreasing.

The reason π is added on the left side and 2π is added on the right side is because the starting interval of all intervals of x when Arccoscosx is decreasing is (π,2π), noting that π is the ending point to the interval of when Arccoscosx is increasing. Also note that it takes a period of π to go from one decreasing interval to the next.

Since Cosx’s domain is [0,π]:

Then, Arccoscosx = ArccoscosX = Arccoscos(x - 2kπ) = x - 2kπ

Then, Arccoscosx = ArccoscosX = Arccoscos(2kπ + 2π - x) = 2kπ + 2π - x

Note that the laws of inequalities was used.

Therefore, there are two possible answers of simplifying Arccoscosx:

**Sketch the following function:**

**y = sinArcsinx**

Begin by simplifying the function’s expression.

Let t = Arcsinx, then if t = Arcsinx, x = sint.

sint = x

Therefore, the graph of y = sinArcsinx looks like y = x, however the domain and range is different.

The domain of the inner function, Arcsinx is [-1,1]. The domain of the entire function simplified, x, is (-∞,∞). However, the inner function’s domain is important to consider, so the final domain is [-1,1].

The range of the entire function is y(Dt), where Dt is the domain of t, or Arcsinx. The domain of Arcsinx is [-1,1]. The range of y(Dt) is y simplified with t’s domain, or x’s range with Arcsinx’s domain, which is still [-1,1].

Therefore, knowing that sinArcsinx is the shape of y = x, with a domain and range of [-1,1], the following is the graph of y = sinArcsinx:

**Differentiate the following function:**

**y = sinArcsinx**

Since sinArcsinx was simplified to x:

However, one can differentiate using the chain rule and simplification:

Let t = Arcsinx, then if t = Arcsinx, x = sint.

**Differentiate the following function:**

Using the chain rule:

**Differentiate the following function:**

Using the chain rule, assuming x is the independent variable:

**Differentiate the following function:**

**f(t) = tArctant**

Using the product rule and chain rule:

**Differentiate the following function:**

**F(x) = (1 + x²)Arctanx**

Using the product rule and the chain rule:

**Differentiate the following function:**

Using the quotient rule and the chain rule:

**Differentiate the following function:**

Using the chain rule:

**Differentiate the following function:**

Using the quotient rule and the chain rule:

**Differentiate the following function:**

Using the subtraction rule and the chain rule:

**Solve the following initial-value problem:**

When asked to solve an initial-value problem, it is asking for the original function of y, given an initial point.

Antidifferentiating y’:

Finding C, when y(0) = 1:

y = Arctanx + C

1 = Arctan0 + C

1 = 0 + C

C = 1

Therefore, y = Arctanx + 1

**Solve the following initial-value problem:**

When asked to solve an initial-value problem, it is asking for the original function of y, given an initial point.

Antidifferentiating y’:

Finding C, when y(½) = 1:

y = Arcsinx + C

1 = Arcsin(½) + C

1 = π/6 + C

C = 1 - π/6

Therefore, y = Arcsinx + 1 - π/6

**Simplifying the following expression:**

**cosArctanx**

Let t = Arctanx, then if t = Arctanx, x = tant.

An identity that relates to cosine and tangent is the following:

Since x = tant:

Therefore:

**Simplify the following expression:**

**sinArctan2x**

Let t = Arctan2x, then if t = Arctan2x, then 2x = tant.

Finding an identity that relates with sine and tangent:

Finding an identity that relates with cosine and tangent:

Since 2x = tant:

Therefore:

**Find f(x) of the following derivative, if f(1) = 0:**

Antidifferentiating f’(x):

Finding C, when f(1) = 0:

Therefore:

**Evaluate the following expression:**

**tanArcsec4**

Let t = Arcsec4, then if t = Arcsec4, sect = 4.

Since sect = 1/cost = r/x, assume r = 4 and x = 1. Using the Pythagorean theorem, y = √15.

Therefore:

tanArcsec4 = tant = sint/cost = (√15/4)/(¼) = √15

**Evaluate the following expression:**

**sin(2Arcsin(3/5))**

Let t = Arcsin(3/5), then if t = Arcsin(3/5), sint = 3/5.

Assume y = 3 and r = 5. Using the Pythagorean theorem, x = 4.

Therefore:

sin(2Arcsin(3/5)) = sin2t = 2sintcost = 2(3/5)(4/5) = 6/5

**Evaluate the following expression:**

**Arccossin(6π/5)**

First, the point of sin(6π/5) must be found before the angle Arccossin(5π/6) is found. Since 6π/5 is not a known angle, it will be sketched using logical reasoning.

Since π + π/5 = 6π/5, 6π/5 is greater than π and creates a π/5 angle between π and the terminal arm of 6π/5. This creates a right triangle, where the vertical leg is labeled t because of the need to find sine’s value, and the horizontal leg is labeled the resulting x value found when using y = t and r = 1 into the Pythagorean theorem.

If y = -t, x = -√(1 - t²), and r = 1, then sin(6π/5) = y/r = -t/1 = -t.

Then Arccossin(6π/5) = Arccos(-t) = some angle θ, therefore:

cosθ = -t

Since cosθ is a negative value and Arccosine’s domain is in between 0 and π, the triangle formed in quadrant III is moved to quadrant II. Additionally, sin(6π/5) was originally -t, but now it’s cosθ’s value needed to be found, so the x and y values change in the moved triangle as well.

Therefore:

Arccossin(6π/5)

–> cosθ = -t

–> θ = π/2 + π/5 = 7π/10

Sorting objects by their string fields using....arctan?

I’m pretty proud of myself for this one.

While at work a few days ago, I was writing some code for a comparator to sort an object by String fields alphabetically. The way it was working, it was about 3 lines of simple ternary operators returning the appropriate integer value for the comparator by checking the result of String.compareTo(), which returns the lexagraphical distance between the two strings passed in. This is because comparators want -1 for “less than”, 0 for “equal”, and 1 for “greater than”, and we need to convert the lexagraphical distance to the appropriate result. It was originally something like this:

int comparison = String.compareTo(o1.getStringField(), o2.getStringField());

int result = comparison != 0 ? 1 : 0;

return comparison;

I thought for a minute about this: what if we could directly map the lexagraphical value to the appropriate comparator value? We’d need a function that took all values greater than or equal to 1 and mapped them to 1, map 0 to 0, and mapped values less than or equal to -1 to -1. I immediately thought of the arctan function.

However, the original wouldn’t work here: the asymptotes are at +/-pi/2 which would mean that any rounding could make 2s or -2s, and we want to stay strictly within the result set of {0,1,-1}. Furthermore, arctan(1) is less than ½, so that would round down to 0.

Thankfully, some simple high school trig saves the day. We want some constant integer c such that c\*arctan(x) had asymptotes at +/-1; this means c is 2/pi. We want constant integer d such that |(2/pi)arctan(dx)| is greater than ½ for all x that are integers greater than or equal to 1; I settled on d=4.

So, how do we use this function f(x) = (2/pi)arctan(4x)? Well, we pass the lexagraphical distance result as x, and then call Math.rint() on the result, which maps to the nearest integer. This is why we wanted the value of f(x) at 1 to be greater than ½ – so it could map to 1 by rounding. The resulting single line of code looks like this:

Return Math.rint((2/Math.PI)*Math.atan(4*String.compareTo(o1.getStringField(), o2.getStringField())));

Of course, I’m not a jackass: I cleared this with the project lead and wrote a comment clearly explaining how this all works, but this is how to use the arctangent function to sort Strings in Java.

- decide what angle
*ϕ*you want to rotate by - select the part of the image with your text in it [1]
- click the button in the top bar that says ‘Resize’
- make sure the ‘Maintain aspect ratio’ box is turned off
- grab a calculator that supports trigonometric functions [2]
- calculate 100*cos(
*ϕ*) and put it in the first box [3] - calculate 100/cos(
*ϕ*) and put it in the second box - put -
*ϕ*in the third box - calculate arctan(sin(
*ϕ*)cos(*ϕ*)) and put it in the fourth box - transform!

[1] note MS Paint does not support layers
so any part of the picture right next to your text will also be rotated

[2] Google or Wolfram Alpha will do it, but make sure to type ‘degrees’ to tell it your angle is in degrees not radians

[3] you can’t put decimals in the MSPaint resize box so just round it to the nearest whole number…

Derivation for the above formulae here. tl;dr: multiply shear and scale matrices together, set equal to rotation matrix, solve equation.

Sorry for being late.

I made two ways of graphical explanation of Euler’s formula.

This explanation is the same as wikipedia.

Another explanation post is here.

Fig 1

The *N-*th power of *a* (complex number) is located as shown, and all triangles are similarity. because the absolute values are multiplied and the arguments are added to yield the polar form of the product. the points are on a logarithmic spiral.

Fig 2

Put *a*=(1+*i θ*/*N*), and take a limit *N*→∞ then the logarithmic spiral approaches the unit circle. The argument of *a* is equal to arctan(*θ*/*N*). This argument is approximately equal to θ/N in case of N>>1.

Fig 3

The *N-*th power of *a* is located as shown.

Fig 4

To sum up, take the limit *N*→∞ then *a^N* approaches cos(*θ*)+*i *sin(*θ*).

Fig 5

Remember the definition of *e^z*, We obtain Euler’s formula, *e^i θ*=cos(*θ*)+*i *sin(*θ*).

‘Memba them .gifs?! Had to design myself some business cards and this one for arctan(z), z∈[0, 2π], ended up being the inspiration for the backs of ‘em, and then when I got home earlier from my near-daily failed attempt at getting shit done for my class at a coffee shop in town, I made my first go at… something? conformal mapping-esque as a concertina card using the same thing but with 15 contours instead of 26 since that’d be a total shitshow. (This isn’t a conformal map, this a contour map of the real/blue and imaginary/red parts, whatever, but it’s around the same level of intricacy probably.) Unfortunately I was kind of sloppy in making my Adobe Illustrator cut pattern because THE INSPIRATION WAS COMING TOO HARD TOO FAST so the fold wasn’t centered, but oh well. I also had to do a lot of post-op manual Xacto assistance because I still haven’t gotten the hang of this machine. First time coloring one of the machine-cut ones, too. It’s curled worse than any of my hand-cut ones, gonna blame the cardstock. I’m about to try and get some of this homework done before I go disappoint my complex teacher with how little I’ve accomplished tomorrow and then I think I’mma finally give doing long strips of Bristol a whirl.

Also that was not the final version of the front of my business cards, that’s an older iteration without my phone number. Also also it makes me super happy that my logo font is Euclid, the contact info is Gravity. Ha. (And sometime between now and when I leave, I’ve gotta get an actually-professional portfolio site up, gah. STRESSIN’. And procrastinating. By doing more art projects. Sorry not sorry.)

like simple trig… is easy but solving trig equations took me fifty goddamn years to master….. arctan(arcsin (5pi/3) - arccos (8pi/5)) is still engraved in my eyelids bc those problems r literally such a pain in the ass

*Re[arctan(z)] and Im[arctanh(z)]*

The angle shown in the first photo corresponds with the front view of one branch of Im[arctanh(z)] while the angle in the last photo corresponds with the front view of one of Re[arctanh(z)]; in other words, the two functions only differ by a 90**°**** **rotation. The flat edges of the painted areas in the center of the top and bottom Plexiglass plates are the location of two branch cuts, where there is a discontinuity. We could make the functions continuous by stacking countably many more copies of the sculpture (or branches, mathematically speaking) on top of each other to construct their Riemann surfaces, as these branch cuts correspond with where one [-π, π] iteration of these multivalued functions ends and another begins, which is easy to see in the middle photo.

Re[arctan(x+iy)]

Something a little different: the contour lines of Re[arctan(z)] in black and Im[arctan(z)] in white, x,y∈[0, 2π], 29 contours. I’m really happy with how this turned out (and with the thinner border), and I’m definitely going to do more of them, but for most of these trig functions the contour lines don’t go all the way to the edge for one part or the other, so I’ve got some further experimenting to do. Probably with less contours. Hopefully in color.

And this one’s for Banj, who was found shot dead in his car last night, who made my favorite weird ass trippy doodle cartoons, who always made me laugh, who was the best 2.5th roommate ever, who was a great guy and a good soul, who was doing something with his life and had people that cared about him, who didn’t deserve to die and die so young just because he was in the wrong place at the wrong time. It’s not fair and it doesn’t make sense. So many people I care about are hurting so badly right now. I loved you to death, bro, and I’ll miss you and treasure your weird doodles always.

Re[arctan(z)] (blue) and Im[arctan(z)] (red) plotted from [0, 2π]

holy shit y’all

So when I was writing up the description for my arctan sculptures last post I had initially written [0, 2π] instead of [-π, π] and then thought about it a little harder and realized I’d never actually looked at any of these trig functions from [0, 2π], all the plots and animations and everything I’d ever done was always [-π, π]. And I looked at what arctan looked like, and I wondered what the hell the [0, 2π] version would look like and kind of imagined it as something that puckered out in the middle and tapered off at the ends. WRONG. NOT EVEN CLOSE, AND WOAH THAT’S COOL. So I am now sitting in a chaise lounge in the fancy hotel this conference is at just trying to hammer out animations of all the ~shape variations~ of all four “trig families” (can I just start calling them that for my own sake? The regular, inverse regular, hyperbolic, and inverse hyperbolic? They lil families? Mom, dad, brother, sister, baby trig functions? I just really like making up my own terms for things to make my life easier?).

Anyway, here’s the first one, and with a color scheme that actually follows the one I’m more or less standardizing for my sculptures: real parts cool colors, imaginary parts warm colors, since the real numbers are ~familiar and comforting and calm~ and complex numbers are (legitimately) more dynamic and full of life and movement herp derp derp. So yeah just finished up coth so I’ll dump that one on you here shortly. Er mer gerd I’ve missed making these animations, going to hopefully actually download some gif making software so I can make them on the drive back since I usually just use gifmaker.me.

WHAT AN EXCITING TURN OF EVENTS IN THE WORLD OF MY MATH ART BULLSHIT

Re[arctan(x+iy)], x,y∈[0,2π]. This was my first one with colored cardstock. (Also the back of my business cards.)

My two kids, totally filthy and in need of more than the computer duster sponge bath I gave them right as the art exhibit opened, both broke some bones in transit and arctan(z)’s bottom plate is totally busted (sin(xy) just lost a bottom corner which fortunately isn’t super noticeable)… but after arriving in Helsinki, Finland, after 18 hours in airplanes and airports and learning that they hadn’t shown up at the baggage claim because they were back in ATLANTA, GEORGIA, USA, I am surprisingly at peace with them having gotten a little busted on their journey to Jyväskylä because I was just so, so glad to see them again.

I still got all my supplies and materials from makin’ them on my ~*~art cart~*~ down in the physics building demo room, I can set them broken bones and get these lil babies all healed up nbd. The hardest part will probably be walking uphill both ways to the art building to laser cut the new plates in Tennessee in August.

Anyway, BABY’S FIRST INTERNATIONAL ART EXHIBIT! How ‘bout that!

Re[arctan( (x+i y)^5 )]