An Introduction To Numerical Analysis (2) Linear Approximation

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 / 2}Ö (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.

Thus: 

1 + x/2 =  1 + 0.02/2  =  1.01

1 +  x/2 =  1 + 0.05/ 2 =  1.025

1 + x/2 = 1 +   0.005/2 =  1.00250

 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