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Month: August 2026

Building the Sine Function Term by Term

In this post we will solve the following basic second-order equation with corresponding initial conditions:

$\displaystyle y'' = -y, \qquad y(0) = 0, \qquad y'(0)=1$

This equation can be solved very easily exactly. The answer is $y(x) = sin(x)$ . Now, let’s pretend that we don’t know how to solve this equation exactly. We introduce a parameter $\epsilon$ and consider the family of equations

$\displaystyle y'' = -\epsilon y$

$\displaystyle y'' +\epsilon y = 0$

and the function $y(x)$ in the form:

$\displaystyle y_{\epsilon}(x) = a_0(x) + a_1(x)\epsilon + a_2(x)\epsilon^2 + a_3(x)\epsilon^3 + ...$

$\displaystyle y_{\epsilon}''(x) = a_0''(x) + a_1''(x)\epsilon + a_2''(x)\epsilon^2 + a_3''(x)\epsilon^3 + ...$

The differential equation becomes (up to three terms):

$\displaystyle y'' +\epsilon y = 0$

$\displaystyle a_0''(x) + a_1''(x)\epsilon + a_2''(x)\epsilon^2 + a_3''(x)\epsilon^3 + \epsilon (a_0(x) + a_1(x)\epsilon + a_2(x)\epsilon^2 + a_3(x)\epsilon^3) = 0$

$\displaystyle a_0''(x) + (a_1''(x) + a_0(x))\epsilon + (a_2''(x) + a_1(x))\epsilon^2 + (a_3''(x) + a_2(x)) \epsilon^3 = 0$

This implies that:

$\displaystyle a_0''(x) = 0 \implies a_0(x) = x$

$\displaystyle a_1''(x) + x = 0 \implies a_1(x) = -\frac{1}{3!}x^3$

$\displaystyle a_2''(x) -\frac{1}{3!}x^3 = 0 \implies a_2(x) = \frac{1}{5!}x^5$

$\displaystyle a_3''(x) +\frac{1}{5!}x^5 = 0 \implies a_3(x) = -\frac{1}{7!}x^7$

and

$\displaystyle y_{\epsilon}(x) = x -\frac{1}{3!}x^3 \epsilon + \frac{1}{5!}x^5 \epsilon^2 - \frac{1}{7!}x^7 \epsilon^3$

Setting $\epsilon =1$ to recover the original problem leads to:

$\displaystyle y_{\epsilon}(x) = x -\frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7$

which corresponds to the first terms of the Taylor series of $sin(x)$ (the exact solution).

August 6, 2026