Mark Frauenfelder at 8:56 am Wed, Nov 16, 2011
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My friend John Edgar Park took the Moustair fad to another level, or five.
That is the most vulgar thing I’ve seen in quite awhile.
What are you going to do when recursion runs wild on you!
I look forward to the infinite zoom animated .gif in the next Hogan Moustair post.
Hulk Head Ad Infinitum. Fantastic :)
Hulkamania exists in the very quantum foam from which everything springs into being, I have always suspected this.
I think Cyriak just got an idea for his next animation…
“Steroids: They’re not just for rage and shriveled testicles anymore”?
In the vein of funny cat ladder problems:
Problem: Hulk Hogan has decided that yo dog he likes his face, so he put his face in his face so he can shave while he shaves. He shrinks his face from ear to ear so that it fits inside his moustache, with the top of the bridge of his nose ending up just underneath his moustache.
Let the height of Hulk’s face from top of nose bridge to bottom of moustache be h, and the width of Hulk’s face from ear to ear be w. Further, let the width of the inside of Hulk’s moustache be some fraction of the width of Hulk’s face, 0 < k < 1.
What is the length L of Hulk’s recursive moustache starting from the bottom of his original moustache?
Solution: The next iteration of Hulk’s face must be multiplied in size by k in order to fit inside his moustache. Thus, the height of the next iteration is hk. The iteration after is again multiplied in size by k, resulting in an additional height of hk^2. Continuing, we get a sum from 1 to n of hk^n.
We know that the infinite sum of k^n from 0 to infinity is 1/(1-k), thus the infinite sum from 1 to n of hk^n is h/(1-k) – h, and so the length of Hulk’s fractal moustache is L = h/(1-k) – h.
Making a substitution u=1/hk results in 1 = huk, which proves that (rather recursively) that the length of Hulk’s recursive moustache is huLk.
Human Centipede 3
Kill it with fire and drown the ashes with unicorn blood!!!
Unicorn blood keeps things alive even when they’re 99% dead already. Turban sold separately.
this image requires a chaser.
The Hasselhoff recursion is much, much worse.
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