MA005: Calculus I

An approximation is most valuable if we also have have some measure of the size of the "error", the distance between the approximate value and the value being approximated. Typically, we will not know the exact value of the error (why not?), but it is useful to know that the error must be less than some number. For example, if one scale gives the weight of a gold pendant as $10.64$ grams with an error less than $.3$ grams $(10.64 \pm.3$ grams) and another scale gives the weight of the same pendant as $10.53$ grams with an error less than $.02$ grams $(10.53 \pm.02$ grams), then we can have more faith in the second approximate weight because of the smaller "error" guarantee. Before finding a guarantee on the size of the error of the linear approximation process, we will check how well the linear approximation process approximates some functions we can compute exactly. Then we will prove one bound on the possible error and state a somewhat stronger bound.

Example 8: Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$. Evaluate $\mathrm{f}(2+\Delta \mathrm{x}), \mathrm{L}(2+\Delta \mathrm{x})$ and $|\mathrm{f}(2+\Delta \mathrm{x})-\mathrm{L}(2+\Delta \mathrm{x})|$ for
                  $\Delta \mathrm{x}=0.1,0.05,0.01,0.001$ and for a general value of $\Delta \mathrm{x}$.

Solution: $\quad \mathrm{f}(2+\Delta \mathrm{x})=(2+\Delta \mathrm{x})^{2}=2^{2}+4 \Delta \mathrm{x}+(\Delta \mathrm{x})^{2}$ and $\mathrm{L}(2+\Delta \mathrm{x})=\mathrm{f}(2)+\mathrm{f}^{\prime}(2) \cdot \Delta \mathrm{x}=2^{2}+4 \cdot \Delta \mathrm{x}$. Then

$\begin{array}{l|l|l|l} \Delta \mathrm{x} & \mathrm{f}(2+\Delta \mathrm{x}) & \mathrm{L}(2+\Delta \mathrm{x}) & |\mathrm{f}(2+\Delta \mathrm{x})-\mathrm{L}(2+\Delta \mathrm{x})| \\ \hline 0.1 & 4.41 & 4.4 & 0.01 \\ 0.05 & 4.2025 & 4.2 & 0.0025 \\ 0.01 & 4.0401 & 4.04 & 0.0001 \\ 0.001 & 4.004001 & 4.004 & 0.000001 \end{array}$

Cutting the value of $\Delta x$ in half makes the error $1 / 4$ as large. Cutting $\Delta x$ to $1 / 10$ as large makes the error $1 / 100$ as large. In general, $|\mathrm{f}(2+\Delta \mathrm{x})-\mathrm{L}(2+\Delta \mathrm{x})|=\left|\left(2^{2}+4 \cdot \Delta \mathrm{x}+(\Delta \mathrm{x})^{2}\right)-\left(2^{2}+4 \cdot \Delta \mathrm{x}\right)\right|=(\Delta \mathrm{x})^{2}$.

This function and error also have a nice geometric interpretation (Fig. 8): $f(x)=x^{2}$ is the area of a square of side $x$ so $f(2+\Delta x)$ is the area of a square of side $2+\Delta x$, and that area is the sum of the pieces with areas $2^{2}, 2 \cdot \Delta x, 2 \cdot \Delta x$, and $(\Delta x)^{2}$. The linear approximation $\mathrm{L}(2+\Delta x)=2^{2}+4 \cdot \Delta x$ to the area of the square includes the 3 largest pieces $2^{2}, 2 \cdot \Delta x$ and $2 \cdot \Delta x$, but it omits the small square with area $(\Delta x)^{2}$ so the approximation is in error by the amount $(\Delta x)^{2}$.

Practice 10: Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$. Evaluate $\mathrm{f}(4+\Delta \mathrm{x}), \mathrm{L}(4+\Delta \mathrm{x})$ and

$|\mathrm{f}(4+\Delta \mathrm{x})-\mathrm{L}(4+\Delta \mathrm{x})|$ for $\Delta \mathrm{x}=0.1,0.05,0.01,0.001$ and for a general value of $\Delta x$. Use Fig. 9 to give a geometric interpretation of $f(4+\Delta x)$, $\mathrm{L}(4+\Delta \mathrm{x})$ and $|\mathrm{f}(4+\Delta \mathrm{x})-\mathrm{L}(4+\Delta \mathrm{x})|$.

In both the example and practice problem, the error $|\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})|$ turned out to be very small, proportional to $(\Delta x)^{2}$, when $\Delta x$ was small. In general, the error approaches $\mathrm{0}$ as $\Delta x$ approaches $\mathrm{0}$.

Theorem :      If $\quad \mathrm{f}(\mathrm{x})$ is differentiable at a and $\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a}) \cdot \Delta \mathrm{x}$

Proof: $\quad|\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{L}(\mathrm{a}+\Delta \mathrm{x})|=\left|\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{a})-\mathrm{f}^{\prime}(\mathrm{a}) \cdot \Delta \mathrm{x}\right|=\left\{\frac{\mathrm{f}(\mathrm{a}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{a})}{\Delta \mathrm{x}}-\mathrm{f}^{\prime}(\mathrm{a})\right\} \cdot \Delta \mathrm{x}.$ But $\mathrm{f}$ is differentiable

at $x=\mathrm{a}$ so $\lim _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}=\mathrm{f}^{\prime}(\mathrm{a})$ and $\lim _{\Delta x \rightarrow 0}\left\{\frac{f(a+\Delta x)-f(a)}{\Delta x}-f^{\prime}(a)\right\}=0$.

Then $\lim _{\Delta x \rightarrow 0}|f(a+\Delta x)-L(a+\Delta x)|=\lim _{\Delta x \rightarrow 0}\left\{\frac{f(a+\Delta x)-f(a)}{\Delta x}-f^{\prime}(a)\right\} \cdot \lim _{\Delta x \rightarrow 0} \Delta x=0 \cdot 0=0$

Not only does the difference $f(a+\Delta x)-L(a+\Delta x)$ approach $\mathrm{0}$, but this difference approaches $\mathrm{0}$ so fast that we can divide it by $\Delta \mathrm{x}$, another quantity approaching $\mathrm{0}$, and the quotient still approaches $\mathrm{0}$.

In the next chapter we can prove that the error of the linear approximation process is proportional to $(\Delta \mathrm{x})^{2}$. For now we just state the result.

Theorem: If $\mathrm{f}$ is differentiable at $\mathrm{a}$ and $\left|\mathrm{f}^{\prime \prime}(\mathrm{x})\right| \leq \mathrm{M}$ for all $\mathrm{x}$ between $\mathrm{a}$ and $\mathrm{a}+\Delta \mathrm{x}$