ONE MORE STEP TOWARD BETTER TOMORROW : Linear Approximations

Tuesday, 26 March 2013

Linear Approximations

This approximation is crucial to many known numerical techniques such as Euler's Method to approximate solutions to ordinary differential equations. The idea to use linear approximations rests in the closeness of the tangent line to the graph of the function around a point.
Let x₀ be in the domain of the function f(x). The equation of the tangent line to the graph of f(x) at the point (x₀,y₀), where y₀ = f(x₀), is

$\begin{displaymath}y - y_0 = f'(x_0) (x-x_0) \cdot\end{displaymath}$

If x₁ is close to x₀, we will write $x_1 = x_0 + \Delta x$ , and we will approximate $f(x_0 + \Delta x)$ by the point (x₁,y₁) on the tangent line given by

$\begin{displaymath}y_1 = y_0 + \Delta x f'(x_0) \cdot\end{displaymath}$

If we write $\Delta y = y_1 - y_0$ , we have

$\begin{displaymath}\Delta y = \Delta x f'(x_0) \cdot\end{displaymath}$

In fact, one way to remember this formula is to write f'(x) as $\displaystyle \frac{dy}{dx}$ and then replace d by $\Delta$ . Recall that, when x is close to x₀, we have

$\begin{displaymath}f(x) \approx f(x_0) + f'(x_0)(x-x_0) \;\cdot\end{displaymath}$

Example. Estimate $\sqrt{9.2}$ .
Let $f(x) = \sqrt{3+x}$ . We have $f(6) = \sqrt{9}= 3$ . Using the above approximation, we get

$\begin{displaymath}f(6.2) \approx f(6) + f'(6) (6.2-6)\end{displaymath}$

We have

$\begin{displaymath}f'(x) = \frac{1}{2} \frac{1}{\sqrt{x+3}} \cdot\end{displaymath}$

So $f'(6) = \displaystyle \frac{1}{6}$ . Hence

$\begin{displaymath}f(6.2) \approx f(6) + f'(6) (6.2-6) = 3.033\end{displaymath}$

or $\sqrt{9.2} \approx 3.033$ . Check with your calculator and you'll see that this is a pretty good approximation for $\sqrt{9.2}$ .Remark. For a function f(x), we define the differential df of f(x) by

$\begin{displaymath}df = f'(x)\, dx \;\cdot\end{displaymath}$

Example. Consider the function y = f(x) = 5x². Let $\Delta x$ be an increment of x. Then, if $\Delta y$ is the resulting increment of y, we have

$\begin{displaymath}\begin{array}{lll} \Delta y &=& f(x+ \Delta x) - f(x)\\ &=&... ...\ &=& 10 x (\Delta x) + 5 (\Delta x )^2\;\cdot\\ \end{array}\end{displaymath}$

On the other hand, we obtain for the differential dy:

$\begin{displaymath}dy = f'(x)\,dx = 10 x\,dx\;.\end{displaymath}$

In this example we are lucky in that we are able to compute $\Delta y$ exactly, but in general this might be impossible. The error in the approximation, the difference between dy(replacing dx by $\Delta x$ ) and $\Delta y$ , is $5 (\Delta x )^2$ , which is small compared to $\Delta x$ .

ONE MORE STEP TOWARD BETTER TOMORROW

Tuesday, 26 March 2013

Linear Approximations

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