Thread: approximating error on taylor poly

1. approximating error on taylor poly

Find the fifth order taylor polynomial for f(x) = 1/x^2 based at x_0 = 1, then find an interval centered at x_0 = 1 in which the approximation error |f(x) - P_5(x)| < 0.01

I found the taylor poly
1 -2(x-1) + 3(x-1)^2 -4(x-1)^3 ... -6(x-1)^5

I don't know how to simply find the approximation. I could try solving for x but that would take way too long.

2. Originally Posted by diroga
I found the taylor poly
1 -2(x-1) + 3(x-1)^2 -4(x-1)^3 ... -6(x-1)^5

I don't know how to simply find the approximation. I could try solving for x but that would take way too long.
The error in an alternating series is always less than the n+1 term

$\displaystyle 7(x-1)^6<\frac{1}{100} \iff (x-1)^6<\frac{1}{700}$

$\displaystyle (x-1)< \left( \frac{1}{700} \right)^{\frac{1}{6}} \approx .336$

3. Hello,

You have an expression of the rest : $\displaystyle R(x)=\frac{1}{5!}\int_{1}^{x}f^{(6)}(t)(x-t)^5dt$

After having calculate this integral we obtain :

$\displaystyle R(x)=\frac{7(x^6-6x^5+15x^4-20x^3+15x^2-6x+1)}{x^8}$

Then you study this function on $\displaystyle [0,2]$ for example.

And we find this interval : $\displaystyle [0,76;1,24]$

Maybe it exists an easier solution.

Edit : Yes, I didn't think to use series...