Extension: Error Analysis in Linear Approximation

Previously, we learned how to use the method of linear approximation to estimate values of functions near a point. Specifically, we found that for a small change in $x$ from $x=a$, denoted by $\Delta x$, $$f(a+\Delta x)\approx L(x)=f(a)+f'(a)\Delta x.$$ So, the error of such an approximation is found by simply taking the difference between the approximation and the actual value: $$E(x)=|f(x)-L(x)|=|f(a+\Delta x)-[f(a)+f'(a)\Delta x]|$$ $$=|[f(a+\Delta x)-f(a)]-f'(a)\Delta x|=|\Delta y-\frac{dy}{dx}\Delta x|.$$ (Notice how all of the differentials appear to cancel!)

Let's look at an example, and calculate the potential error when estimating $f(x)=x^2$. We have $$f(a+\Delta x)=(a+\Delta x)^2=a^2+2a\Delta x+(\Delta x)^2$$ and $$L(a+\Delta x)=f(a)+f'(a)\Delta x=a^2+(2a)\Delta x.$$ Therefore, $$E(x)=|f(a+\Delta x)-L(a+\Delta x)|\\=|[a^2+2a\Delta x+(\Delta x)^2]-[a^2+2a\Delta x]|\\=|\Delta x|^2.$$ First, observe that as $\Delta x\to 0$, the error term also approaches $0$, which is what we expect. In addition, for small values of $\Delta x$, the quantity $|\Delta x|^2$ is significantly smaller than the quantity $\Delta x$, which appears in the formula for linear approximation. So, this tells us that the error term is relatively small.

Now you try!

We approach the problem very similarly. We have $$f(a+\Delta x)=(a+\Delta x)^3=a^3+3a^2\Delta x+3a(\Delta x)^2+(\Delta x)^3$$ and $$L(a+\Delta x)=f(a)+f'(a)\Delta x=a^3+(3a^2)\Delta x.$$ Therefore, $$E(x)=|f(a+\Delta x)-L(a+\Delta x)|\\=|[a^3+3a^2\Delta x+3a(\Delta x)^2+(\Delta x)^3]-[a^3+(3a^2)\Delta x]|\\=|3a(\Delta x)^2+(\Delta x)^3|.$$

Again, as $\Delta x\to 0$, the error term also approaches $0$. As before, this is expected.

(Video) Linear Approximation and Error

by Math Videos from Heather

Here's a video by Math Videos from Heather explaining how to find the error of a linear approximation.