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

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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0The 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?

TuringTest
 2 years ago
Best 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?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0reading 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0Never.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.
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