08Glinearapproximations


 * Linear approximation **

Euler's Rule

Consider the definition of derivative studied in the previous chapter. math . \qquad f'(x) = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {f(x+h) - f(x)}{h} math

If h is very small the derivative can approximated. math . \qquad f'(x)\approx\dfrac {f(x+h) - f(x)}{h} math

Rearranging this equation we obtain ** Euler's rule **, (an approximation formula). math . \qquad f(x+h)\approx\; f(x) + hf'(x) math

Alternatively, the difference between the values is math . \qquad f(x+h) - f(x) \approx hf'(x) math

This can be rewritten using (delta) to mean a small change in x (or y) math . \qquad \delta{y}=\delta{x}\dfrac{dy}{dx} \qquad \textit{an alternate form of Euler's Rule} math


 * Example 1 **

Use Euler's approximation formula with f(x) = x 2 to find an approximate value for (3.02) 2.

__**Solution:**__

math . \qquad f(x+h) \approx \; f(x) + hf'(x) math where math \\ . \qquad f(x) = x^2 \quad \text{ so } \quad f'(x)=2x \\. \\ . \qquad \text{and } x=3 \quad \text{ and } \quad h = 0.02 math

math \\ . \qquad f(3.02) \approx f(3) + 0.02\times f'(3) \\. \\ . \qquad f(3.02) \approx 9 + 0.02\times 6 \\. \\ . \qquad f(3.02) \approx 9.12 math


 * Example 2 **

Use Euler's approximation formula math \text{to find an approximate value for}\;\sqrt[3]{7}. math

__Solution:__ math . \qquad f(x+h) \approx \; f(x) + hf'(x) math where math . \qquad f(x)=\sqrt[3]{x} \quad \text{ so } \quad f'(x)=\dfrac{1}{3}x^{-\frac{2}{3}} math

math \\ . \qquad \text{We know }\sqrt[3]{8}=2, \\. \\ . \qquad \text{therefore use } x = 8 \;\; \text{ and } \;\; h=-1 math

math \\ . \qquad f(7)\approx f(8) -1 \times f'(8) \\. \\ . \qquad f(7)\approx 2 -1 \times \dfrac{1}{3 \times \left( \sqrt[3]{8}\right)^2 } \\. \\ . \qquad f(7)\approx 2-\dfrac{1}{12} \\. \\ . \qquad f(7)\approx 1 \frac{11}{12} math

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