Obfuscation Challenge #1 (Basic -> Advanced) May 28, 2015 obfuscate ob·fus·cate \ˈäb-fə-ˌskāt; äb-ˈfəs-ˌkāt, əb-\: to make (something) more difficult to understand Challenge: Can you create the following graph? Rules: You may not repeat any previous answer. Share this:Click to share on Twitter (Opens in new window)Click to share on Facebook (Opens in new window)Like this:Like Loading... Related
24 thoughts on “Obfuscation Challenge #1 (Basic -> Advanced)”
y = x
It looks like y=x
And it is! Can you come up with a different function that has the same graph?
Ok, here’s my second answer!
parametric equation: x(t) = t, y(t) = t
y=2(1/2(x)-1/2) + 1. Not very creative, but definitely not the same as y=x.
-pi/4<t<pi/4 (one period) but since Desmos doesn't allow pi for t I chose -.8<=t<=.8 (covers slightly more than one full period).
Here’s another! https://www.desmos.com/calculator/zqtypvrcja
y = tan(arctan(x))
Would that create holes at x = (2n+1)(pi/2), for n integer?
Here’s one: https://www.desmos.com/calculator/1xtzaono7e
y = floor(x) + mod(x,1)
Mine is a bit /mean/ since I did leave off a few people’s responses. https://www.desmos.com/calculator/x5qs01vz9a
Very creative. Here’s my not so creative answer: https://www.desmos.com/calculator/v5d2n2b2jj
Gauntlet thrown down! calculus, statistics, summation, PI notation, taylor series, etc.
Theta = pi/4 in polar form
Got another here: https://www.desmos.com/calculator/azwovxy77y
Nothing like a little differentiation and trig substitution!
Obfuscated y=x, I took the graph of the normal curve, shifted it right, reduced the standard deviation, took the derivative of the curve and found the equation for the line tangent to the curve at the point below. I think it works.
No doubt it is y=x …perfect!
Nicely done Laila!
Would that yield a removable discontinuity at (0,0)?
How about y=d(.5x^2+6)/dx ?
Great! First calculus answer??