E X: Fill & Download for Free

Suppose [math]f(x)[/math] is one of the undiscovered functions you’re talking about that is its own derivative. The function [math]e^x[/math] satisfies this property too, but it has already been discovered. By the product rule of differentiation,[math]\dfrac{d}{dx}e^{-x}f(x) = -e^{-x}f(x)+e^{-x}f^\prime(x)=-e^{-x}f(x)+e^{-x}f(x)=0.[/math]This means that [math]e^{-x}f(x)[/math] is a constant [math]C[/math], implying that [math]f(x)=Ce^x[/math].This confirms that the only functions that are equal to their own derivatives are the already well-known functions of the form [math]Ce^x[/math], [math]C\in\mathbb{R}[/math].

One day, [math]e^x[/math] sees [math]x^2[/math] running down the street in a panic."What's wrong?" asks [math]e^x[/math]."There's a Differential Operator in town!" yells [math]x^2[/math]. "If I run into him too many times, I'll disappear!""Don't worry," responds [math]e^x[/math]. "I'll go have a chat with him. No, don't worry about me -- he can't hurt me. After all, I'm [math]e^x[/math]."So [math]e^x[/math] walks down the street to the Differential Operator. "My friend tells me you're a Differential Operator," [math]e^x[/math] says pompously. "Well, I'm [math]e^x[/math].""Pleased to meet you, [math]e^x[/math]," says the Differential Operator. "I'm [math]d/dt[/math]."

Let's say that [math]X[/math][math][/math] is the number of coin flips that you need in order to get two heads in a row. We want to calculate [math]E(X)[/math].We can condition E(X) on whatever our first flip is. I will let [math] E(X|H) [/math] denote the number of remaining flips I need to get two heads in a row conditional on the fact that I have already rolled a head. I let [math] E(X|T) [/math] denote the same for tails.By first step conditioning, we have[math]E(X) = \frac{1}{2}(1 + E(X|H)) + \frac{1}{2}(1 + E(X|T)) [/math]Now, I note that [math]E(X|T) = E(X) [/math] since if I flipped a tail on my first try, I have made no progress towards my two heads goal and I have to start over.Additionally, now note that I can also split up [math]E(X|H)[/math] by first step conditioning[math]E(X|H)=\frac{1}{2}(1+E(X|HH))+ \frac{1}{2}[/math][math](1+E(X|HT)) [/math]Similar to above, [math] E(X|HT)) = E(X)[/math]Also, note that [math]E(X|HH) = 0[/math], since I need no more additional flips.So now, I am left with the following equations:[math]E(X) = \frac{1}{2}(1 + E(X|H)) + \frac{1}{2}(1 + E(X)) [/math][math]E(X|H) = \frac{1}{2}(1 + 0) + \frac{1}{2}(1 + E(X)) [/math]And this is just a simple systems of equations, that when solved gives me[math]E(X) = 6[/math]