Sentence examples for max we can from inspiring English sources

Since b j - i ≥ dist i max, we can have bj-i ≥ disti.

Then, based on b j ≥ r j max, we can conclude bj ≥ rj.

Given a Q max, we can have different values for (P_{a}^{text {max}}) depending on (textbf {O}left (frac {1}{Q_{text {max}}}right)).

Since β > a max, we can choose a constant k > 1 satisfying ( 1 − a min β ) k + 1 β ∑ i = 1 n b i max ≤ k and 1 β ∑ i = 1 n b i min 1 + k q < 1.

When g H S ∗ g≤Q≤ Qmax, we can first express h as (boldsymbol {h}=eta _{boldsymbol {h}} vec {boldsymbol {g}}+xi _{boldsymbol {h}} vec {boldsymbol {g}}_{perp }).

Suppose the result is (p(1),p(2)…p(max)); we can use the following equations to calculate l min VM i ) and l max VM i ): l_{text{min}}=left{ begin{aligned} p text{floor}(m)+1 -1.5*Q & quad {m notin Z} (p(m)+p(m+1))/2-1.5*Q & quad text{other} end{aligned} right.

Although g is not continuously differentiable due to the max function, we can use the subgradient instead [21], where γ n is obtained from Equation 14.

Hence, exploiting the associativity of the max operator, we can perform the two types of pooling operations in one go, as illustrated on the right hand side of Fig. 4.

By substituting (46) and (47) into P r 1 t - opt = P max t and P a t - opt = P max t, respectively, we can derive the corresponding demarcation point T ab = Td 1where P r 1 t - opt achieves its maximal value, and can also derive the corresponding T ab = Td 2where P a t - opt achieves its maximal value.

By setting (22a), (22b), and (22c) to be P max t, respectively, we can obtain TTWR = Td 1when P a t - opt = P max t, TTWR = Td 2when P b t - opt = P max t, and TTWR = Td 3when P r t - opt = P max t.

where P{r>A max} is the probability that r, the amplitude of x 1 t), is larger than the clipping threshold A max. Thus, we can write begin{array}{*{20}l} P left{r>A_{max}right} &= int_{A_{max}}^{R_{max}} f_{x_{1}}left(rright) dr, end{array} (6).