Thursday, September 18, 2008

why derivatives ?

To find an approximation of \sqrt[3]{25} one can do as follows.

  1. Consider the function f(x)= x^{1/3}.\, Hence, the problem is reduced to finding the value of f(25).
  2. We have
    f\ '(x)= 1/3x^{-2/3}.
  3. According to linear approximation
    f(25) \approx f(27) + f\ '(27)(25 - 27) = 3 - 2/27.
  4. The result, 2.926, lies fairly close to the actual value 2.924…
Read some interesting and basic mathematics.. http://en.wikipedia.org/wiki/Linear_approximation
at
Labels:

No comments:

Post a Comment