Problems that become easier in a more general form

Solution 1:

Consider the following integral $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx$. All of our attempts at finding an anti-derivative fail because the antiderivative isn't expressable in terms of elementary functions.

Now consider the more general integral $f(y) = \displaystyle\int_{0}^{1}\dfrac{x^y-1}{\ln x}\,dx$.

We can differentiate with respect to $y$ and evaluate the resulting integral as follows:

$f'(y) = \displaystyle\int_{0}^{1}\dfrac{d}{dy}\left[\dfrac{x^y-1}{\ln x}\right]\,dx = \int_{0}^{1}x^y\,dx = \left[\dfrac{x^{y+1}}{y+1}\right]_{0}^{1} = \dfrac{1}{y+1}$.

Since $f'(y) = \dfrac{1}{y+1}$, we have $f(y) = \ln(y+1)+C$ for some constant $C$.

Trivially, $f(0) = \displaystyle\int_{0}^{1}\dfrac{x^0-1}{\ln x}\,dx = \int_{0}^{1}0\,dx = 0$. Hence $C = 0$, and thus, $f(y) = \ln(y+1)$.

Therefore, our original integral is $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx = f(7) = \ln 8$.

This technique of generalizing an integral by introducing a parameter and differentiating w.r.t. that parameter is known as Feynman Integration.

Solution 2:

I recall something like this coming up when evaluating certain summations. For example, consider:

$$ \sum_{n=0}^{\infty} {n \over 2^n} $$

We can generalize this by letting $f(x) = \sum_{n=0}^{\infty} nx^n$, so:

$$ \begin{align} {f(x) \over x} &= \sum_{n=0}^{\infty} nx^{n-1} \\ &= {d \over dx} \sum_{n=0}^{\infty} x^n \\ &= {d \over dx} {1 \over {1-x}} = {1 \over (x-1)^2} \end{align} $$

Therefore,

$$ f(x) = {x \over (x-1)^2} $$

The solution to the original problem is $f({1 \over 2}) = 2$.

Solution 3:

George Polya's book How to Solve It calls this phenomenon "The Inventor's Paradox": "The more ambitious plan may have more chances of success." The book gives several examples, including the following.

1) Consider the problem: "A straight line and a regular octahedron are given in position. Find a plane that passes through the given line and bisects the volume of the given octahedron." If we generalize this to "a straight line and a solid with a center of symmetry are given in position..." it becomes very easy. (The plane goes through the center of symmetry and the line.)

The book also gives other examples of the Inventor's Paradox, but "more ambitious" is not always the same as "more general." Consider: "Prove that $1^3 + 2^3 + 3^3 + ... + n^3$ is a perfect square." Polya shows that it is easier to prove (by mathematical induction) that "$1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ...+ n)^2$". This is more ambitious but is not more general.

ADDED LATER:

The web page Generalizations in Mathematics gives many similar examples. It even gets into the difference between "more ambitious" and "more general."

Solution 4:

The solution to the Monty Hall problem

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, follows the fixed protocol of opening another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

becomes more obvious when you generalize it to an $N$-door problem with the host opening $N-2$ doors. For $N\gg3$ most people's intuition revolts against staying with the original choice.