ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 19

Sunday, 3 August 2014

CHAPTER 19 - Errors and Approximations

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable. The theory behind it is quite simple: From the chapter on differentiation, we know that

$\mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\Delta y}}{{\Delta x}} = \dfrac{{dy}}{{dx}} = f'\left( x \right)$

For small $\Delta x$ , we can therefore approximate $\Delta y\,{\rm{as}}\,f'\left( x \right)\Delta x$ . This is all there is to it!

Suppose we have to calculate ${(4.016)^2}$ .

We let

	$y = {x^2}.{x_0} = 4\,\,\,{\rm{and}}\,\,{y_0} = 16$
	$y' = 2x,\,\,\Delta x = 0.016$
	$\Rightarrow \,\,\,\, \Delta y = f'\left( x \right) \cdot \Delta x$
	$= {\left. {2x} \right\|_{{x_0} = 4}} \times 0.016$
	$8 \times 0.016$
	$=0.128$
	$\Rightarrow \,\,\,\, y = {y_0} + \Delta y = 16.128$

Example: 1

Find the value of ${\left( {8.01} \right)^{4/3}} + {\left( {8.01} \right)^2}$

Solution: 1

Let $y = f\left( x \right) = {x^{4/3}} + {x^2}$

Let ${x_0} = 8\,\,\,{\rm{so}}\,\,{\rm{that}}\,\,{y_0} = 16 + 64 = 80$

	$\Delta x = 0.01$
	$\Rightarrow \,\,\,\, \Delta y = {\left. {f'\left( x \right)} \right\|_{x = {x_0}}} \times \Delta x$
	${\left. { = \left( {\dfrac{4}{3}{x^{1/3}} + 2x} \right)} \right\|_{{x_0} = 8}} \times \Delta x$
	$= \left( {\dfrac{8}{3} + 16} \right) \times 0.01$
	$= \dfrac{{0.56}}{3}$
	$= 0.1867$
	$\Rightarrow \,\,\,\, {y_0} = {y_0} + \Delta y$
	$= 80.1867$

Sunday, 3 August 2014

CHAPTER 19 - Errors and Approximations

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