linear approximation of sqrt(82)

Question

anonymous · Answer

I understand the concept of linearization and linear approximation. However unsure of calculating a simple single term such as the above. Would I do the following:

Using the linearization law of that:

f(x) ~~ l(x) = f(a) - f'(a)(x-a)

In the case above we would LET a = 80.

Thus

f(x) = sqrt(x)

phi · Answer

you use the first 2 terms of a taylor series expansion http://en.wikipedia.org/wiki/Taylor_series#Definition f(x) $\approx$f(a) + f'(a) (x-a)/1! in your case, you would pick the nearest number that has an integer square root (rather than an irrational number) i.e. pick a=81, so f(a)=9 $$ f'(x) = \frac{1}{2}x^{-\frac{1}{2}} $$ evaluated at x=81, you get f'(a)= 1/18 your approximation is $$f(x) ≈ 9 + \frac{x-81}{18} $$ or $$f(x) ≈ 4.5 + \frac{x}{18} $$