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TuringTest
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Mean Value Theorem: is there a typo on the third line?
http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partcmeanvaluetheoremantiderivativesanddifferentialequations/session34introductiontothemeanvaluetheorem/MIT18_01SCF10_ex34prb.pdf
 one year ago
 one year ago
TuringTest Group Title
Mean Value Theorem: is there a typo on the third line? http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partcmeanvaluetheoremantiderivativesanddifferentialequations/session34introductiontothemeanvaluetheorem/MIT18_01SCF10_ex34prb.pdf
 one year ago
 one year ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.0
The paper is drawing a parallel between linear approximations so it talks about switching out \(f'(a)\) for \(f'(c)\) for some specific yet undetermined c but the last formula on the page notes that this idea extended leads to something like a quadratic approximation, which relates to Taylor series\[f(b)=[f(a)+f'(a)(ba)]+\frac{f''(c)}2(ba)^2\]should it not be\[f(b)=[f(a)+f'(c)(ba)]+\frac{f''(c)}2(ba)^2\]or is the whole point using \(f'(a)\)to keep it related to linear approximations?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[f(b)\approx f(a)+f'(a)(ba)\]near \(x=a\), whereas MVT states that\[f(b)=f(a)+f'(c)(ba)\]for som \(a<c<b\) is the whole point using \(f'(a)\)to keep it related to linear approximations?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
reading it again I'm thinking the answer to my own question is "no it's not a typo" (since it says the part in brackets is \(exactly\) the linear approximation, which it is with f'(a) ) but if anyone has any more insight let me know
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Never.mind, I think the answer is that it is not a typo. It says the "next term" in the Taylor series will approximate error in an nth degree approximation. I suppose only the next term in the Taylor series would require the c to get an exact equality.
 one year ago
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