Exercise 1, § 3

Functions of a Real Variable, VI, 296

Show that if $f^{(2p+1)}(t)$ is monotone on the interval $[x, x+n+1]$ then the remainder $\mathrm{T}_p(x, n)$ in the Euler-Maclaurin formula VI, p. 288 (formula (2)) has the same sign as the term

\frac{1}{(2p+2)!}\, b_{2p+2}\, \left( f^{(2p+1)}(x+n+1) - f^{(2p+1)}(x) \right)

and has absolute value at most equal to that of this term.

If one assumes that $f^{(2p+2)}$ is continuous (but of any sign), show that one has

\left| \mathrm{T}_p(x, n) \right| \leqslant \frac{1}{(2p+2)!} \left| b_{2p+2} \right| \left( \left| f^{(2p+1)}(x+n+1) - f^{(2p+1)}(x) \right| + \int_x^{x+n+1} \left| f^{(2p+2)}(t) \right| dt \right).

Solution

No solution has been written yet.