Introduction to Approximations and Taylor Series

Once you have started learning the rules for solving derivatives, etc, what is the purpose of learning approximations? What needs to be approximated?

From the University of Aberdeen Math Department : (A PDF version of the document is available)

Approximations are very important in mathematics, particularly in its applications. There are very many problems that we do not know how to solve exactly but for which it is comparatively easy to get an approximate answer. These are not ideal from a purely mathematical point of view but in practical work there is very little difference between knowing that the answer to a problem is $ \sqrt{{2}}$ and knowing that it is approximately 1.414214 to 6 decimal places.

For example, there are no formulas that will allow you to write down the exact solutions of even such apparently simple equations as x8 +3x5 – 2x + 1 = 0, sin x = x cos x, ex = 5x. With these it is approximation or nothing.

There is a related problem that is just as important and which we will consider later. How do you actually work out the values of functions like the trig functions or the exponential? If you want to know e2.3 or sin 2.1 then you press a button on your calculator. But how does the calculator know how to do it? In a strict sense it doesn’t. All it knows is how to obtain the value to the level of accuracy required by the display of the calculator. So this is once more an approximation problem.

I am now going to look at this approximation from a rather different point of view. Rather than regarding it as a way of approximating the value of the derivative I am going to assume that we know the derivative and really want to find f (a + h).

Why would we want to do that? Imagine we wanted to find out the square root of 9.1…The square root of 9.0 is easy to figure out, three. The square root 9.1 isn’t so easy, so use an approximation. These approximations work best when the h value is small. From he Linear Approximation section:

The trick here is to notice that 9.1 is quite close to 9 and we certainly know $ \sqrt{{9}}$ without using a calculator. In other words, we can write 9.1 = 9 + h, where h = 0.1. Let f (x) = $ \sqrt{{x}}$. Then f’(x) = $\displaystyle {\frac{{1}}{{2\sqrt{x}}}}$. So the linear approximation formula becomes

$\displaystyle \sqrt{{x+h}}$ $\displaystyle \approx$ $\displaystyle \sqrt{{x}}$ + $\displaystyle {\frac{{h}}{{2\sqrt{x}}}}$

This will be `true’ for any positive value of x so long as h is small enough. In our case we want to put x = 9 and h = 0.1. Then we get

$\displaystyle \sqrt{{9.1}}$ $\displaystyle \approx$ $\displaystyle \sqrt{{9}}$ + $\displaystyle {\frac{{0.1}}{{2\sqrt{9}}}}$ = 3 + $\displaystyle {\frac{{0.1}}{{6}}}$ = 3.01667

Note: there may be an error on one of the pages. The author wrote that d/dx (1+x)/(1-x) is 2/(1+x)^2 . However, isn’t the answer 2/(1-x)^2, using the quotient rule?

Where the input to the first f(x) term and the derivatives are set x=zero in the Taylor series formula, that’s a Maclaurin series. See Taylor Series (Wikipedia).

More later…

Internal tag: math


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