Is $$\sum \limits_{t=a}^{b} f(t) \leq \int \limits_{a-c}^{b+c} f(t)dt$$
if $f(t) \geq 0$ and $f(t)$ is monotonic ?!
I saw on Wikipedia a similar result but i want to generalize it for c some real number that is bigger or equal 1
summation bound by integration?
0
$\begingroup$
inequality
approximation
2 Answers
3
If $c$ is bigger or equal to $1$ and $f\ge 0$ is non-dreasing, this is true: Note that \begin{align*} \sum_{i=a}^b f(i) &= \sum_{i=a}^b \int_i^{i+1} f(i)\, dt\\ &\le \sum_{i=a}^b \int_i^{i+1} f(t)\, dt & \text{due to monotonicity}\\ &= \int_a^{b+1} f(t)\, dt\\ &\le \int_{a-c}^{b+c} f(t)\, dt & \text{$f \ge 0$} \end{align*} If $c < 1$, this need not to be true, note that for $$ f\colon \mathbf R \to \mathbf R, \qquad t\mapsto \begin{cases} 0 & t < a \\ 1 & t \ge a \end{cases} $$ we have $$ \sum_{i=a}^b f(i) = b-a+1, \quad \int_{a-c}^{b+c} f(t)\, dt = b-a+c $$
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I need a little bit of more information, but I'll suppose you have a partition between $a$ and $b$ of integer numbers. Besides $c\geq 1$. If that is true, the sum in left can be seen as the lower Riemann sum of $$I=\int_a^{b+1} \! f(x) \, \mathrm{d}x$$ when f is increasing, and the upper sum for $$J=\int_{a-1}^{b} \! f(x) \, \mathrm{d}x$$ when is decreasing. As f is no-negative, both I, J are bounded by the integral of right.