quadratic approximations

[kjbohn]

1) Give the general cubic and quatric approximations for a function f near a.
2) Use a quadratic approximation to compute as the lim approches 0. (1-cos(x))/x.
3) Using a quartic approximation to approximate 930^1/2 determine which is better starting at point a, 30^2 or a,31^2.

I know that the linear approximation for a function near a point a is given by L(x)= f(a)+f'(a)(x-a). And a quadratic approximation is given by Q(x)=f(a)+f'(a)(x-a)+1/2f''(a)(x-a)^2. But I have no idea where to go from there. Please help me.
Thank you

 

[galactus]

2) Use a quadratic approximation to compute as the lim approches 0. (1-cos(x))/x.

\(\displaystyle \lim_{x\to 0}\frac{1-cos(x)}{x}\)

Use \(\displaystyle f(x)\approx f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2}(x-x_{0})^{2}\)

Which becomes \(\displaystyle f(x)\approx f(0)+f'(0)x+\frac{f''(0)}{2}x^{2}\)

Since \(\displaystyle f(x)=cos(x)\), then \(\displaystyle f(0)=1, \;\ f'(x)=-sin(x), \;\ f'(0)=0, \;\ f''(x)=-cos(x), \;\ f''(0)=-1\)

So we get, upon entering into the quad approx, \(\displaystyle cos(x)\approx 1-\frac{x^{2}}{2}\)

So, the limit \(\displaystyle \lim_{x\to 0}\frac{1-(1-\frac{x^{2}}{2})}{x}=\lim_{x\to 0}\frac{x}{2}=0\)

See how to do it now?.