Problems that become easier in a more general form

When solving a problem, we often look at some special cases first, then try to work our way up to the general case.

It would be interesting to see some counterexamples to this mental process, i.e. problems that become easier when you formulate them in a more general (or ambitious) form.


Recently someone asked for the solution of $a,b,c$ such that $\frac{a}{b+c} = \frac{b}{a+c} = \frac{c}{a+b} (=t).$

Someone suggested writing this down as a system of linear equations in terms of $t$ and solving for $a,b,c$. It turns out that either (i) $a=b=c$ or (ii) $a+b+c=0$.

Solution (i) is obvious from looking at the problem, but (ii) was not apparent to me until I solved the system of equations.

Then I wondered how this would generalize to more variables, and wrote the problem as:
\frac{x_i}{\sum x – x_i} = \frac{x_j}{\sum x – x_j} \quad \forall i,j\in1,2,\dots,n

Looking at this formulation, both solutions became immediately evident without the need for linear algebra (for (ii), set $\sum x=0$ so that each denominator cancels out with its numerator).


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.

Source : Link , Question Author : MGA , Answer Author : Joel Reyes Noche

Leave a Comment