$f(x)=ln(x-\sqrt[3]{x^3-1})$ is been calculated at $x=35$ and is store in $8$ significant numbers.
1.Find the size of the error
2.Suggest a better way to achieve higher precision and find the absolute error
How should I approach it?
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gbox
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Taylor polynomial? – Martín Vacas Vignolo Dec 29 '16 at 19:02
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@vvnitram how will it help if I can only store 8 significant numbers? – gbox Dec 29 '16 at 19:16
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1Same question, without an answer: should this be a duplicate? – shardulc Dec 29 '16 at 22:04
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It would probably helpful if you computed the values, once with rounding in every step and once with the full accuracy of your computer/calculator. Then think about applying Taylor polynomials or binomial identities, and compute again with rounding in every computation step. – Lutz Lehmann Dec 30 '16 at 08:40
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@shardulc : Not exactly duplicate, since there the rounding is a mix of floating and fixed point, while here it is a clean floating point rounding. – Lutz Lehmann Dec 30 '16 at 08:41
1 Answers
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As per Taylor $$x-\sqrt[3]{x^3-1}=\frac1{3·x^2}+\frac1{9·x^5}+\frac{5}{81·x^8}+O(x^{-11})$$ the difference has 3 zeros after the decimal point. Both terms in the difference have 2 digits before the decimal point and are thus rounded to 6 digits after, the difference can only have a maximum of 3 correct digits. This can result in a relative error of $10^{-2}$, which translates into an absolute error of that magnitude in the logarithm.
Or seen in another way, the rounding error of the root term is of size $35·10^{-8}$. In the difference, this error remains but is now relative to a base value of about $\frac1{3·35^2}\sim\frac1{3000}$, which translates in the logarithm to an absolute error of the same magnitude of about $3·35^3·10^{-8}\simeq 0.0013$.
Even that is still rather pessimistic in view of the actual difference of the direct evaluation
log(2.7200000e-04)=-8.2097085to the more accurate valuelog(2.7211096e-04)=-8.2093006.You can use the binomial identity $$ A-B=\frac{A^3-B^3}{A^2+AB+B^2} $$ using $A=x$ and $B=\sqrt[3]{x^3-1}$.
You can also continue to use the Taylor expansion to get $$ \ln(x-\sqrt[3]{x^3-1}) =\ln(3·x^2)+\ln\left(1+\frac1{3·x^3}+O(x^{-6})\right) \\=\ln(3·x^2)+\frac1{3·x^3}+O(x^{-6}) $$
Lutz Lehmann
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