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Linearization

April 10th, 2009 |

Author: Merrin

In this section, we will play some games with the definition of the derivative in order to make some approximate calculations. We have as our definition of the derivative.

$\cfrac{d}{dx} f(x)=\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x)-f(x)}{\Delta x}$

What happens if we multiply both sides of the equation by

$\Delta x$

${\mathop{\lim }\limits_{\Delta x\to 0} \Delta xf'(x)=\mathop{\lim }\limits_{\Delta x\to 0} f(x+\Delta x)-f(x)} \\ \\ \\ {\mathop{\lim }\limits_{\Delta x\to 0} f(x+\Delta x)=f(x)+\mathop{\lim }\limits_{\Delta x\to 0} \Delta xf'(x)}$

This is exact, but if we relax the condition that we take the limit as Delta x goes to zero then we have a pretty good approximation. Then we can write what is called the linearization.

$f(x+\Delta x)\approx f(x)+f'(x)\Delta x$

This approximation is good as long as Delta x is small. Let us try an example.

Example 1. Estimate the square root of 27.

Solution 1. We will use the linearization formula. We pick a function

${f(x)=\sqrt{x} } \\ {f'(x)=\cfrac{1}{2\sqrt{x} } }$

We choose Delta x to be 2 and choose x to be equal to 25. We know how to take the square root of 25 because it is a perfect square and 25 is also near 27 so we can keep our Delta x small relative to the square root of 25. Our linearization formula states

$\sqrt{27} \approx \sqrt{25} +\cfrac{1}{2\sqrt{25} } 2=5+\cfrac{1}{5} =5.2$

The actual value from our calculator gives 5.196152. Not bad!

Later we will learn how to make even higher order corrections beyond linearization which will allow us to make calculations with even greater accuracy. There are even formulas for the error, so the business of calculation becomes scientific. Soon you will become a human calculator with the help of calculus.

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