The use of Taylor series as a means of approximation to a function's value was previously examined in terms of the challenge problem worked out (See Solutions to Pt. 1) . We saw, for example, how to obtain the value of Ö 5 using an approximation to 4th order terms, e.g.
P 4 = f(x) + f ’(x) + f ’’(x)/2! + f ’’’(x)/3! + f iv (x)/4! =
2 + ½ x -1/2 + (-1/4 x -3/2 )/2! + (3/8 x -5/2 ) /3! + (-15/16 x -7/2)/ 4!
There are many cases, however, where such higher orders aren't needed, especially if the root is likely to be 0 or near 0. In these cases a linear approximation is often employed, limited to the first two terms of a Taylor series for the function.
Consider obtaining an approximation for:
f(x)
= Ö (1 + x)
at x = 0
In this case the applicable (truncated) Taylor series will be, if we evaluate at a = 0:
f(a) + f ’(a) × (x – a)
For which f '(x) = ½ (1 + x) -1/2
For the given values we have:
f(a) + f ’(a) × (x – a) = Ö (1 + 0) + ½ (1 + 0)(x – 0) -1/2
= 1 + x / 2Ö (0 + 1) = 1 + x/2
The graph of the function vs. its linearization is shown below:
As may be seen, the intersection of the two functions occurs at the x = 0, y= 0 point. We can get increasing approximation accuracy simply by refining the numerator of the 2nd term.
The last result accurate to 5 decimals, while the next to last is at 3 decimals and the first at 2 decimals. It follows that in general we may write for this approximation:
Ö (1 + x) = 1 + x/2
Suggested Problems:
1) Find the linear approximation to:
f(x) = Ö (1 + x) at x = 3
2) Show that in general:
(1 + x)k » 1 + kx
For any k, provided x » 0
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