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Above is the taylor series centred at a including the lagrange remainder. Can t be equal to a or b?

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  • $\begingroup$ In Michael Spivak's book: Calculus, it is proved in Chapter 20 Theorem 4 (I'm referring to 3rd edition) that $t$ can be chosen to lie in the open interval $(a,b)$. $\endgroup$
    – peek-a-boo
    Jun 9, 2019 at 13:09

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You can take $t$ to lie on the open interval $(a,b)$.

Reference: wolfram mathworld

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