Making a problem easier by making it harder

In the oral exam for my PhD, my advisor asked me a question about a differential equation. I don’t recall the question, but I remember the interaction that followed.

I was stuck, and my advisor countered by saying “Let me ask you a harder question.” I was still stuck, and so he said “Let me ask you an even harder question.” Then I got it.

By “harder” he meant “more general.” He started with a concrete problem, then made it progressively more abstract until I recognized it. His follow-up questions were logically harder but psychologically easier.

This incident came to mind when I ran across an example in Lawrence Evans’ control theory course notes. He uses the example to illustrate what he calls an example of mathematical wisdom:

It is sometimes easier to solve a problem by embedding it within a larger class of problems and then solving the larger class all at once.

The problem is to evaluate the integral of the sinc function:

\int_0^\infty \frac{\sin(x)}{x}\, dx

He does so by introducing the more general problem of evaluating the function

I(\alpha) = \int_0^\infty \exp(-\alpha x) \frac{\sin(x)}{x}\, dx

which reduces to the sinc integral when α = 0.

We can find the derivative of I(α) by differentiating under the integral sign and integrating by parts twice.

I'(\alpha) &=& \int_0^\infty \frac{\partial}{\partial \alpha} \exp(-\alpha x) \frac{\sin(x)}{x}\, dx \\ &=& \int_0^\infty \exp(-\alpha x) \sin(x) \, dx \\ &=& - \frac{1}{1 + \alpha^2}

Therefore

I(\alpha) = - \arctan(\alpha) + C

As α goes to infinity, I(α) goes to zero, and so C = π/2 and I(0) = π/2.

Incidentally, note that instead of computing an integral in order to solve a differential equation as one often does, we introduced a differential equation in order to compute an integral.

8 thoughts on “Making a problem easier by making it harder

  1. Thanks. I didn’t know that term, but it’s a handy one.

    The idea of “liberating constraints” is similar: becoming more creative by giving yourself fewer options. Maybe they’re dual concepts. The inventor’s paradox weakens your hypotheses and liberating constraints strengthen your conclusion.

  2. There was a beautiful example of this on 538.com’s The Riddler a while back.

    https://fivethirtyeight.com/features/can-you-bake-the-optimal-cake/

    Working out the math for the 3-layer cake is a horrible mess of algebra, but the riddle ends with two questions, ” What percentage of the cone’s volume does the cake fill?”, and “What if he had asked for an N-layer cake?”. If you concentrate on these you get a short, neat inductive formula with just a fraction of the algebraic pain.

    I particularly liked the way those questions were stuck on at the end, as if they were mere afterthoughts, but they’re actually the keys to the elegant solution,

  3. It is true that making a problem more general can be an effective problem solving technique. But there’s something else going on in some of these examples.

    When my advisor asked me a more general question, he was also giving me information. He knew the answer, and was steering me toward it by focusing my attention. The same with the cake problem. When someone generalizes the problem for you, they’re giving you a hint.

  4. I remember a trick like this for evaluating gamma function integrals too.

    btw. Looks like you dropped a sign on the second line of the I'(\alpha) calculation.

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