(a) Show that 1+kx is the local linearization of (1+x)^k near x=0

Question

a) Show that 1+kx is the local linearization of (1+x)^k near x=0

(b) Someone claims that the square root of 1.1 is about 1.05. Without using a calculator, do you think that this estimate is about right? 

(c) Is the actual number above or below 1.05? 

Answer 1

f(x) = (1 + x)^k

 

 

a) Taylor series for (1 + x)^k at x = 0:

 

 

f(x) = f(0) + f'(0)x + ½f"(0)x² + ...

 

 

f(x) = 1^k + k(1^k-1)x + ½k(k -1)(1^k-2)x² + ... =

 

 

1 + kx + ½k(k - 1)x² + ...

 

 

Keeping up to the linear term, f(x) ≅ 1 + kx

 

 

b) Use the result above to estimate √1.1:

 

√1.1 = (1 + 0.1)^½ ≅ 1 + (½)(0.1) = 1.05  with x = 0.1 ant k = ½.

 

The closer x is to zero, the better the approximation.

 

 

c)  The next term in the Taylor series is ½k(k-1)x².  Each succeeding term in the Taylor series will be at least an order of magnitude less than the previous since x = 0.1.

 

For x = 0.1 and k = ½, the quadratic term will be negative and this will decrease the approximation for √1.1 when it is included as the third term in the Taylor expansion of f(x).

 

We expect the actual value of √1.1 to be less than 1.05.