Re: Induction with a hard start

José Carlos Santos wrote: 
Could someone please provide an example of a proof by induction in which
proving the statement for n = 1 is clearly harder than going from n to
n + 1?

In the development of the theory of calculus one usually derives the basic rules of differentiation before giving formulas for the derivative of specific functions. So in this process one comes to the point where the chain rule, etc. are known, but the derivative of the trigonometric functions are unknown. Finding the n'th order derivative of sin(x) provides an example of an induction with a hard start:

Statement: The n'th order derivative of sin(x) is sin(x+(n*pi)/2).

This is obviously equivalent to the n'th derivative of sin(x) being sin(x), cos(x), -sin(x), or -cos(x) depending on whether n is congruent to 0, 1, 2, or 3 mod 4, respectively.

When proving this statement the induction step is easy. We are allowed to use trigonometric formulas like cos(x)=sin(x+pi/2), the chain rule, and the statement for n=1, namely [sin(x)]'=cos(x).

However, to prove the initial statement one must use the definition of the derivative (as a limit) to show that [sin(x)]'=cos(x). This involves direct computations of
lim_{h->0} sin(h)/h = 1
and
lim_{h->1} (cos(h)-1)/h = 0.
This is harder.

--------
Tore August Kro
.