ca.analysis and odes - Interesting applications (in pure mathematics) of first-year calculus

In number theory, here are four applications of techniques or results in first-year calculus.

(1) Finding equations of tangent lines by first-semester calculus methods lets us add points on elliptic curves using the Weierstrass equation for the curve. This is more algebraic geometry than number theory, so I'll add that the methods show if the Weierstrass equation has rational coefficients then the sum of two rational points is again a rational point.

(2) The recursion in Newton's method from differential calculus is the basic idea behind Hensel's lemma in $p$-adic analysis (or, more simply, lifting solutions of congruences from modulus $p$ to modulus $p^k$ for all $k geq 1$).

(3) The infinitude of the primes can be derived from the divergence of the harmonic series (the zeta-function at 1), which is based on a bound involving the definition of the natural logarithm as an integral.

(4) Unique factorization in the Gaussian integers can be derived from the Leibniz formula
$$
frac{pi}{4} = 1 - frac{1}{3} + frac{1}{5} - frac{1}{7} + frac{1}{9} - cdots = sum_{n geq 0} frac{(-1)^n}{2n+1}
$$
by interpreting it as a case of Dirichlet's class number formula $2pi h/(wsqrt{|D|}) = L(1,chi_D)$ for $chi_D$ the primitive quadratic character associated to ${mathbf Q}(sqrt{D})$ where $D$ is a negative fundamental discriminant, $h$ is the class number of ${mathbf Q}(sqrt{D})$ and $w$ is the number of roots of unity in ${mathbf Q}(sqrt{D})$. Taking $D = -4$ turns the left side into $2pi h/(4sqrt{4}) = (pi/4)h$, so the Leibniz formula is equivalent to $h = 1$, which is another way of saying $mathbf Z[i]$ is a PID or equivalently (for Dedekind domains) a UFD.

Here are two more applications, not in number theory directly.

(5) Gerry Edgar mentions in his answer Niven's proof of the irrationality of $pi$, which is available in Spivak's calculus book. The same ideas imply irrationality of $e^a$ for every positive integer $a$, which in turns easily implies irrationality of $e^r$ for nonzero rational $r$ and thus also irrationality of $log r$ for positive rational $r not= 1$. The calculus fact in the proof of irrationality of the numbers $e^a$ is that for all positive integers $n$ the polynomial
$$
frac{x^n(1-x)^n}{n!}
$$
and all of its higher derivatives take integer values at $0$ and $1$. That implies a certain expression involving a definite integral is a positive integer, and then with the fundamental theorem of calculus that same expression turns out to be less than 1 for large $n$ (where "large" depends on the hypothetical denominator of a rational formula for $e^a$), and that is a contradiction.

(6) Prove that if $f$ is a smooth function (= infinitely differentiable) on the real line and $f(0) = 0$ then $f(x) = xg(x)$ where $g$ is a smooth function on the real line. There is no difficulty in defining what $g(x)$ has to be if it exists at all, namely
$$
g(x) = begin{cases}
f(x)/x, & text{ if } x not= 0, \
f'(0), & text{ if } x = 0.
end{cases}
$$
And easily the function defined this way is continuous on the real line and satisfies $f(x) = xg(x)$. But why is this function smooth at $x = 0$ (smoothness away from $x = 0$ is easy)? You can try to do it using progressively messier formulas for higher derivatives of $g$ at 0 by taking limits, but a much slicker technique is to use the fundamental theorem of calculus to write
$$
f(x) = f(x) - f(0) = int_0^x f'(t),dt = xint_0^1 f'(xu),du,
$$
which leads to a different formula for $g(x)$ that doesn't involve cases:
$$
g(x) = int_0^1 f'(xu),du.
$$
If you're willing to accept differentiation under the integral sign (maybe that's not in the first-year calculus curriculum, but we used first-year calculus to get the slick formula for $g(x)$) then the right side is easily checked to be a smooth function of $x$ from $f$ being smooth.