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Thresholding to minimize round-off errors (hubbard)



The 2019 Stack Overflow Developer Survey Results Are InLaw of Cosines for very acute angles, round-off errorHow to round off timestamp in milliseconds to nearest seconds?How to compute Fourier coefficients using a cubic spline-corrected FFT?Prof claimed that $arcsin x + arccos x$ does not always round to 90º under the same number of significant figures.Round-off: cross- vs. dot-productsNumerical analysis 1 round off errorsWhy is 'catastrophic cancellation' called so?Round Off Error AnalysisComputing Absolute and Relative Error Using Double PrecisionRound off to decimals










0












$begingroup$


I'm working through Hubbard's Vector Calculus text, and I'm a little fuzzy concerning a detail of one example:



enter image description here



He later asks you to analyze where the errors occur in formula 2.1.12, and I see that
$$
1-2cdot10^10=1-20000000000=-19999999999=-20000000000=-2cdot10^10
$$

due to computing to 10 significant digits--the "error" is that the 1 essentially disappears in the subtraction.



My question is how the 1 becomes $-10^10$ from the second matrix to the third. It would seem that, out to 10 significant figures, we would have
$$
1-10^10=1-10000000000=-9999999999neq-10^10.
$$

Where am I going wrong?










share|cite|improve this question











$endgroup$











  • $begingroup$
    yeah, Technically, that should be $-9999999999$. But does that one change suddenly rescue the calculation and give you the correct result? Or does it just change the amount of error?
    $endgroup$
    – Paul Sinclair
    Mar 24 at 2:35















0












$begingroup$


I'm working through Hubbard's Vector Calculus text, and I'm a little fuzzy concerning a detail of one example:



enter image description here



He later asks you to analyze where the errors occur in formula 2.1.12, and I see that
$$
1-2cdot10^10=1-20000000000=-19999999999=-20000000000=-2cdot10^10
$$

due to computing to 10 significant digits--the "error" is that the 1 essentially disappears in the subtraction.



My question is how the 1 becomes $-10^10$ from the second matrix to the third. It would seem that, out to 10 significant figures, we would have
$$
1-10^10=1-10000000000=-9999999999neq-10^10.
$$

Where am I going wrong?










share|cite|improve this question











$endgroup$











  • $begingroup$
    yeah, Technically, that should be $-9999999999$. But does that one change suddenly rescue the calculation and give you the correct result? Or does it just change the amount of error?
    $endgroup$
    – Paul Sinclair
    Mar 24 at 2:35













0












0








0





$begingroup$


I'm working through Hubbard's Vector Calculus text, and I'm a little fuzzy concerning a detail of one example:



enter image description here



He later asks you to analyze where the errors occur in formula 2.1.12, and I see that
$$
1-2cdot10^10=1-20000000000=-19999999999=-20000000000=-2cdot10^10
$$

due to computing to 10 significant digits--the "error" is that the 1 essentially disappears in the subtraction.



My question is how the 1 becomes $-10^10$ from the second matrix to the third. It would seem that, out to 10 significant figures, we would have
$$
1-10^10=1-10000000000=-9999999999neq-10^10.
$$

Where am I going wrong?










share|cite|improve this question











$endgroup$




I'm working through Hubbard's Vector Calculus text, and I'm a little fuzzy concerning a detail of one example:



enter image description here



He later asks you to analyze where the errors occur in formula 2.1.12, and I see that
$$
1-2cdot10^10=1-20000000000=-19999999999=-20000000000=-2cdot10^10
$$

due to computing to 10 significant digits--the "error" is that the 1 essentially disappears in the subtraction.



My question is how the 1 becomes $-10^10$ from the second matrix to the third. It would seem that, out to 10 significant figures, we would have
$$
1-10^10=1-10000000000=-9999999999neq-10^10.
$$

Where am I going wrong?







algebra-precalculus numerical-methods rounding-error






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 16:34







interrogative

















asked Mar 23 at 15:57









interrogativeinterrogative

265111




265111











  • $begingroup$
    yeah, Technically, that should be $-9999999999$. But does that one change suddenly rescue the calculation and give you the correct result? Or does it just change the amount of error?
    $endgroup$
    – Paul Sinclair
    Mar 24 at 2:35
















  • $begingroup$
    yeah, Technically, that should be $-9999999999$. But does that one change suddenly rescue the calculation and give you the correct result? Or does it just change the amount of error?
    $endgroup$
    – Paul Sinclair
    Mar 24 at 2:35















$begingroup$
yeah, Technically, that should be $-9999999999$. But does that one change suddenly rescue the calculation and give you the correct result? Or does it just change the amount of error?
$endgroup$
– Paul Sinclair
Mar 24 at 2:35




$begingroup$
yeah, Technically, that should be $-9999999999$. But does that one change suddenly rescue the calculation and give you the correct result? Or does it just change the amount of error?
$endgroup$
– Paul Sinclair
Mar 24 at 2:35










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