A model of happiness and three oddly-equivalent definitions

Usually, people are happy (H > 0) when current reality (f(t)) exceeds their expectations, calibrated by the recent past (f(t - x)), and sad (H < 0) when reality subceeds expectation. That is, to a first approximation, H ∝ f(t) - f(t - x), for all t and some x. It would be most fair (symmetric) to consider all x > 0, but weight them differently. More recent times calibrate one's expectations more, according to, say, e^-x. Thus H(t) = ∫_-∞^t ds e^{s - t} (f(t) - f(s)).

That might be a good formula for happiness. My study in that direction ends here. It gets more interesting when seen as a purely mathematical object.

The conclusions are largely the same, just simpler, if we instead work with the related H(t) = ∫_-∞^t ds e^{s - t} f(s).

Some values of that integral for various f:

Notice:

• all derivatives of 1 are 0
• the derivative of t is 1, then all further derivatives are 0
• the derivative of t² is 2t, next 2, and then infinite 0s
• all derivatives of e^t are e^t
• derivatives of cos(t) alternate sine and cosine, positive and negative

In the first three more finite cases, the integral transform gives an alternating sum, f(t) - f'(t) + f''(t). An infinite-series extension — f(t) - f'(t) + f''(t) - f'''(t) + ... — explains the last two, for if 1 - 1 + 1 - 1 + ... has any limit, it would be 1/2.

That is, this infinite integral is equivalent to an infinite series of derivatives. To prove it more rigorously, induce it from repeated integration-by-parts. Pick dv = ds e^{s - t} and u = f(s). Then each step preserves e^{s - t} as-is and differentiates f(s), reversing the sign.

But wait! There's more. That integration-by-parts shows that the indefinite integral ∫ds e^{s - t} f(s) = e^{s - t} (f(t) - f'(t) + f''(t) - f'''(t) + ...) + C. So if we can force C = 0, the infinite definite integral equals the infinite alternating series equals an indefinite integral, followed by a simple operation: (∫ds e^{s - t} f(s)) / e^{s - t}.

You know where else we divide indefinite integrals by exponentials? First-order linear ODEs. In particular, y' + P(t) y = Q(t) is solved with y = (∫dt uQ(x)) / u, where u = e^{∫dt P(t)}.

Plugging in expressions above to "unsolve" the ODE, we get y' + y = f(t). Among the three objects from earlier, we can most readily show that the derivative series satisfies that equation. Differentiating the series reverses the sign on each term, such that y' + y would cancel to 0, but for the initial f(t).

§ tl;dr

The following are equivalent:

• infinite definite integral, ∫_-∞^t ds e^{s - t} f(s)
• alternating derivative series, f(t) - f'(t) + f''(t) - f'''(t) + ...
• transformed indefinite integral, (∫ds e^{s - t} f(s)) / e^{s - t}
• ODE solution, y' + y = f(t)