Edit: As others have pointed out, this is not the best strategy.
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Nice problem! The best strategy seems to be to mix the red clay with the blue clay in small infinitesimal steps. Every bit of red clay then becomes as cold as possible, meaning that as much energy as possible is transferred to the blue clay.
Here’s what we get with two steps:
Mix 1⁄2 red at 100 degrees + 1 blue at 0 degrees ⇒ 33.33 degrees.
Now remove the red clay, and add the other half.
1⁄2 red at 100 degrees + 1 blue at 33.33 degrees ⇒ 55.55 degrees.
Now let’s compute this with n steps. So at each step we add a 1/n fraction of the red clay to the blue clay, then remove it. Let the temperature of the blue clay after k steps be Tk.
Initial: T0=0
Adding a 1/n piece of red clay to the blue clay:
Tk+1=(Tk+100∗1n)/(1+1n)=nTk+100n+1
To simplify our expression, let r be the ratio: nn+1.
Tk+1=r(Tk+100n)
Unrolling our definition gives a geometric series:
Tn=100rn(1+r+...+rn)
Replacing the geometric series with its sum:
Tn=100rn⋅1−rn+11−r
Simplifying, rn(1−r)=nn+1n(1−1n+1)=nn+1nn+1=1
So Tn=100r(1−rn+1).
Now rn+1=(nn+1)n+1=(1−1n+1)n+1. As n approaches infinity, it’s well-known that this approaches 1/e.
Final temperature:
Tn=100(1−1/e) = 63.2 degrees.
As a final note, the LaTeX support for LessWrong is the best I’ve seen anywhere. My thanks to the team!
Edit: As others have pointed out, this is not the best strategy.
___________________________
Nice problem! The best strategy seems to be to mix the red clay with the blue clay in small infinitesimal steps. Every bit of red clay then becomes as cold as possible, meaning that as much energy as possible is transferred to the blue clay.
Here’s what we get with two steps:
Mix 1⁄2 red at 100 degrees + 1 blue at 0 degrees ⇒ 33.33 degrees.
Now remove the red clay, and add the other half.
1⁄2 red at 100 degrees + 1 blue at 33.33 degrees ⇒ 55.55 degrees.
Now let’s compute this with n steps. So at each step we add a 1/n fraction of the red clay to the blue clay, then remove it. Let the temperature of the blue clay after k steps be Tk.
Initial: T0=0
Adding a 1/n piece of red clay to the blue clay:
Tk+1=(Tk+100∗1n)/(1+1n)=nTk+100n+1
To simplify our expression, let r be the ratio: nn+1.
Tk+1=r(Tk+100n)
Unrolling our definition gives a geometric series:
Tn=100rn(1+r+...+rn)
Replacing the geometric series with its sum:
Tn=100rn⋅1−rn+11−r
Simplifying, rn(1−r)=nn+1n(1−1n+1)=nn+1nn+1=1
So Tn=100r(1−rn+1).
Now rn+1=(nn+1)n+1=(1−1n+1)n+1. As n approaches infinity, it’s well-known that this approaches 1/e.
Final temperature:
Tn=100(1−1/e) = 63.2 degrees.
As a final note, the LaTeX support for LessWrong is the best I’ve seen anywhere. My thanks to the team!