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I hope that no one will give me negative votes because I already asked the question in the English usage and meaning community, but no one there was able to tell me the meaning and, by the way, they advised me to ask here!

QUESTION: If we substitute this infinitesimal matrix A into orthogonality condition $A^{t}A = 1$ defined in chapter 1, we will get $(1 + x A')^{t}(1 + A'J) = (1 + x A'^{t})(1 + x A') = 1$, which is same as $\quad x A'^{t} = -x A'\quad$ UP TO first order of $x$.

David G. Stork
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Silvinha
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1 Answers1

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The use of “up to” can be confusing because this expression is used with two quite distinct meanings.

In everyday language, and often also in mathematics, it specifies the upper limit of a range: “An $n\times n$ matrix can have up to $n$ distinct eigenvalues”, that is, it could have less than $n$, or $n$, but not more than $n$ eigenvalues.

However, in mathematics “up to” can also have a different meaning, something like “disregarding”. For instance, “An indefinite integral is only determined up to an additive constant” means that it is essentially determined, but only if we disregard an additive constant. “The angles and side lengths determine a triangle up to translations, rotations and reflections” means that they essentially determine the triangle, but only if we disregard that we can translate, rotate and reflect it. “The prime factorization in a unique factorization domain is unique up to multiplication by a unit” means that the prime factorization is essentially unique, but only if we disregard that we can multiply it by a unit.

Your example is confusing for non-native speakers because “up to” is used in both senses in the context of orders of approximation. In this case, it is being used in the first sense, specifying the upper end of a range, i.e. the two equations are the same if you compare them to zeroth order (the constant terms $1$ cancel) and if you compare them to first order (the two first-order terms have been carried along), but not if you go beyond first order and compare them to second order (the second-order term has been omitted).

But in a very similar formulation, also regarding orders of approximation, “up to” can be used with the second meaning: “This equation is correct up to quadratic terms.” can mean “... if we disregard quadratic terms”, instead of “... up to and including quadratic terms”.

So the equation $\mathrm e^x=1+x$ is correct up to first order (it includes up to linear terms), and it is correct up to quadratic terms, with two different meanings of “up to”.

joriki
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  • I wonder if you meant to link to something else for "first order" – J. W. Tanner Apr 28 '20 at 00:39
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    @J.W.Tanner: I didn't, actually, it was an (apparently failed) attempt at humour :-) I removed it. – joriki Apr 28 '20 at 00:46
  • @joriki Are you able to provide one more example about last meaning of "up to" explained in the equation $e^x=1+x$. Can you provide another example to clarify? – Silvinha Apr 28 '20 at 00:47
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    @VanyushaPavlova: If you search for "up to small terms", you'll find examples where "up to" is used in the second meaning. For instance, from here: "The form factors listed above correspond, up to small terms which are negligible near the peak, to those by Bardin et al. used in the code.", meaning, "disregarding small terms". You'll find examples of "up to" used in the first meaning (upper end of range) in the context of orders of approximation if you search for "expanded up to quadratic terms". – joriki Apr 28 '20 at 00:58