Been working quite a bit on problems where you approximate the value using taylor polynomials and estimating the error. So, I just want to check here that I'm starting to understand it.

"Use second order Taylor polynomials for the given function about the point specified to approximate the indicated value. Estimate the error, and write the smallest interval you can be sure contains the value.

$ f(x) = x^{-1} $ about 1; approximate $ \frac 1 {1.02} $

I start with differentiating $f(x)$ and get

$ f'(x) = \frac {-1} {x^{2}} $

$f''(x) = \frac 2 {x^3} $

$ f'''(x) = \frac {-6} {x^4} $

And after using $f(1)$ I get: $ 1 - 1(x-1) + 1(x-1)^2$ as the function to approximate. When substituting x with $1.02$ I get that $ \frac 1 {1.02} $ almost equals $0.9804$.

Calculating the error $E$, I take the absolute value of: $\left(\frac{f'''(c)}{3!}\right) \cdot (1,02-1)^3$ and when using $c=1$ (since I want the biggest possible error) I get that $ E = \frac 8 {1 000 000} $.

This says that: $ 0.984 - \frac 8 {1 000 000} \leq \frac 1 {1.02} \leq 0.984 + \frac 8 {1 000 000} $.

Is this a correct answer? I've struggled quite a lot on these problems so want to make sure I'm starting to get them now.

  • $\begingroup$ There's something wrong here. You say $f(x)= x^{1/2},$ but that's clearly not the function you're differentiating. It looks to me like you mean $f(x)=x^{-1},$ but then you should have $f'(x) = \frac{-1}{x^2}.$ $\endgroup$ – saulspatz Feb 11 '18 at 19:40
  • $\begingroup$ You're right. I wrote the wrong $f(x)$ and $f'(x)$, should be correct now. Had two tasks infront of me and mixed them together in the beginning. Sorry about that. $\endgroup$ – gbgult Feb 11 '18 at 19:43

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