Hints for the First Question of Homework 1

Let’s apply what we learned in Calculus 1 to the first question of this homework. Given a function $f(x)=x^3$ and $x_0=100$, we want to estimate the relative error of $f(x_0+\mathrm{\Delta }x)$, given that the relative error $\frac{\mathrm{\Delta }x}{x_0}$ is less than $1\mathrm{\%}$. What we need to do here is to estimate

By expanding $f(x_0+\mathrm{\Delta }x)$, we obtain $x_0^3+3x_0^2\mathrm{\Delta }x+3x_0\mathrm{\Delta }{x}^2+\mathrm{\Delta }{x}^3$. If we ignore the terms involving $\mathrm{\Delta }{x}^2$ and $\mathrm{\Delta }{x}^3$, the equation simplifies to $x_0^3+3x_0^2\mathrm{\Delta }x$. This represents the linear approximation of $f$ at $x_0$. Therefore, the relative error of $f$ is

The concepts we learn in Calculus 1 can be generalized to higher dimensions. Suppose we need to calculate the relative error of $f(x,y,z)$, given relative errors for $x$, $y$, and $z$. If we closely follow the definition of the relative error of $f$ and ignore all nonlinear terms of $\mathrm{\Delta }x$, $\mathrm{\Delta }y$, and $\mathrm{\Delta }z$, then we obtain

Then, you will use $\frac{\mathrm{\Delta }x}{x_0}$, $\frac{\mathrm{\Delta }y}{y_0}$, and $\frac{\mathrm{\Delta }z}{z_0}$ to estimate the relative error of $f.$