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# 2011 Calculus BC free response #6d

Lagrange error bound for Taylor Polynomial approximation. Created by Sal Khan.

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• I assume this may be something not required for Mastery of Integral Calculus at Khan Academy as there were no practice problems or Mastery Challenges on this subject, just a proof this subject given in the video series on Taylor Series. Why are there not practice problems on this subject? Why wouldn't this be required for mastery?
• Khan Academy is an on-going project, so there are not yet Mastery Challenges and/or practice problems available for every topic.

But the ability to work with Taylor series is standard calculus curriculum and is needed for applied and advanced level mathematics.
• Is it possible to evaluate M without looking at the graph?
• i still do not understand why the M=40, when M could also equal 80 or 100 or etc. according to this logic. Why can the M be its set to it corresponding value?
• We need a value that leads to a result smaller than 1 / 3000.
Having M = 80 or 100 may lead to a larger result, so the problem may not be solved.
i.e. result > 1 / 3000

M = 40 is nice, because we (1) are sure, (2) has a result smaller than 1 / 3000.
(1 vote)
• Why do you have to know the n+1th derivative of a function to bound the error on the of the approximation? I don't understand the relationship, not even after watching Sal's proof videos.
(1 vote)
• because the n+1th derivative of an n-degree polynomial is 0. If y = x then dy/dx = 1 ... so dy^2/dx = 0 because the derivative of a constant is 0. You can know that's true without knowing anything else... so it gets used to prove the error bound.
(1 vote)
• At 4: 40, why does b have to be greater than a?
(1 vote)
• at around , is it also correct to let M=30 to solve the problem? Or is letting M=40 more correct?
(1 vote)
• Could you also use the alternating series error bound?