- #1
burak100
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upper bound of taylor!
[itex]f(x)[/itex] is two times diff. function on [itex](0, \infty)[/itex] . [itex]\lim\limits_{x\rightarrow \infty}f(x) = 0[/itex] satisfy.
[itex]M=\sup\limits_{x>0}\vert f^{\prime \prime} (x) \vert[/itex] satisfy
. for each integer [itex]L[/itex] ,
[itex]g(L) = \sup\limits_{x\geq L} \vert f(x) \vert[/itex], and [itex]h(L) = \sup\limits_{x\geq L} \vert f^{\prime}(x) \vert[/itex]. for any [itex] \delta > 0[/itex], SHOW
[itex]h(L) \leq \dfrac{2}{\delta} g(L) + \dfrac{\delta}{2}M[/itex].
please helppppp...
[itex]f(x)[/itex] is two times diff. function on [itex](0, \infty)[/itex] . [itex]\lim\limits_{x\rightarrow \infty}f(x) = 0[/itex] satisfy.
[itex]M=\sup\limits_{x>0}\vert f^{\prime \prime} (x) \vert[/itex] satisfy
. for each integer [itex]L[/itex] ,
[itex]g(L) = \sup\limits_{x\geq L} \vert f(x) \vert[/itex], and [itex]h(L) = \sup\limits_{x\geq L} \vert f^{\prime}(x) \vert[/itex]. for any [itex] \delta > 0[/itex], SHOW
[itex]h(L) \leq \dfrac{2}{\delta} g(L) + \dfrac{\delta}{2}M[/itex].
please helppppp...
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