Solutions To Numerical Analysis Problems (2)

1)Find the linear approximation to:

 f(x) =  Ö (1 + x) at x = 3

Solution:

 We evaluate the first two terms of the Taylor series for f at a = 3:

f(a) + f ’(a) × (x – a) 

For which f '(x) =   ½ (1 + x) ^-1/2

f(3) = 2

f '(3) =   ½ (1 + 3) ^-1/2

=     ½  (1/Ö (4))  =   ½ (½)  =  1/4

Then: f(x) =  2 + 1/4  (x - 3) = 2 + x/4 - 3/4

= 5/4  + x/4

At x = 3.2 we obtain:

Ö (1 + x)  =  Ö (1 + 3.2) »  5/4  + 3.2/4

=  1.250 + 0.800 = 2.050  

Which differs from the true value: 

Ö 4.2  »  2.04939   

By less than one one thousandth

2)  Show that in general:

(1 +  x)^k  » 1 +  kx

For any k, provided x  »  0

Solution:

Check at two different values for k:

Rem: Approximation is good for any x »  0

a) k = -1 and we will get (-x) instead of x, so:

(1 +  x) ^k  =   (1 +  x) ^-1 

 1/ (x - 1) = (1 +  x)^-1  »  1 + (-1) (-x) =  1 + x

Try x= 0.01, then:

 (1 +  x)^-1   = 1.01

(1 + kx) = 1.01

But if larger x is tried, e.g. x = 0.1 we see:

(1 +  x)^-1   = 1.11   vs. 1.1 

b) We use k = 1/3  and   5 x⁴   instead of x:

(1 +  x) ^1/3  =    ³Ö( 1 +  5 x⁴ )  =  (1 +  5 x⁴) ^1/3

»  1 + 1/3 (5 x⁴) »  1 + 1/3 (5 x⁴)  »  1 + 5 x⁴/ 3

Try x= 0.50, then:

(1 +  x) ^1/3  =   0.883

(1 + kx) =  1 + 1/3 (5 x⁴) »  0.896

But if smaller x is tried, e.g. x = 0.01, we get equal values (1)