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The equalation $f=\ln({x-\sqrt[3]{x^3-1}})$ is calculated for $x=35$. The root is calculated by taking 8 significiant digits (after the decimal point). What is the error in calculating $f$? find at least one solution which gives better approximation and give approximation for the absolute error in each situation.

I think that there is pervasive error created by loss of significance error.I know that pervasive absolute error is approximated to be $$E\thickapprox f'(x^*)(x-x^*)$$ while the relative error is approximated to be $$Rel(f(x))\thickapprox\frac{x\cdot f'(x)\cdot Rel(f(x))}{f(x)}$$ My problem is I don't know how to applicate these formulas: I don't know how to find $x$ and $x^*$. Furthermore, There is here additionally to the pervasive error also loss of significance that I don't know how to approximate. About a better solution: I think that a btter approximation can be achieved by writing $$\ln(x-\sqrt[3]{x^3-1})=\ln\left(\frac{(x-\sqrt[3]{x^3-1})(x^2+x\cdot\sqrt[3]{x^3-1}+(x^3-1)^{\frac 2 3})}{x^2+x\cdot\sqrt[3]{x^3-1}+(x^3-1)^{\frac 2 3}}\right)$$ $$=\ln{\left(\frac{1}{x^2+x\cdot\sqrt[3]{x^3-1}+(x^3-1)^{\frac 2 3}}\right)}$$ which gives a better approximation. Also here I can't find $x$ and $x^*$ and hence I'm not able to calculate the errors (both absolute and relative). How can I do so?

EDIT:in the lecture we defined pervasive error in the following way: If a calculation is done with numbers which are result of roundoff/chopping/approximation(a.k.a taylor) the result will be influenced from these errors (we calculated $f(x^*)$ instead of $f(x)$). Assuming f' is continious and exists in $[a,b]$ , using lagrange theorm we get $f(x)-f(x^*)=f\prime(c)\cdot(x-x^*)$. if $x^*$ and $x$ are very close the absolute error is $E\thickapprox f\prime(x^*)(x-x^*)$. That is the definition of pervasive error that the lecturer gave.

  • Not quite sure what you mean by "pervasive" error - do you mean roundoff, as in the internal representation in a computer? – Ron Gordon Jul 16 '13 at 16:49
  • I am not sure. See my edit. –  Jul 16 '13 at 17:11
  • Well, what other vaue could $x^*$ be than one arrived at in error? You are tasked with evaluating a function that is subject to loss of significance error - this is due to internal roundoff, as well as error in the binary representation of certain numbers in some cases. – Ron Gordon Jul 16 '13 at 18:01

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