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< a n < a + 1. {\displaystyle \varepsilon =1\Rightarrow \forall n>N(1):a-1<a_{n}<a+1.} Обозначим M = max { a 1 , a 2 , … , a N , a + 1 } ⇒ ∀ n a n ⩽ M...

mat[i][n]=0; for(j=0;j<n;j++) mat[i][n] += mat[i][j]; } /*найдём строку с максимальной суммой*/ for(i=0;i<n;i++) if(mat[i][n]>mat[max][n]) max=i; printf("\n...

max 0 ≤ i ≤ N | y i | ≤ max { | u a | , | U b | } + ( b − a ) 2 8 max 1 ≤ i ≤ N − 1 | f i | ( 4 ) {\displaystyle \max {0\leq i\leq N}|y_{i}|\leq \max\{|u_{a}|...

\forall n>N_{j}\quad \forall z\in {\overline {u}}_{j}\quad \left|f_{n}^{(k)}(z)-f^{(k)}(z)\right|<\varepsilon } Положим N = max j N j {\textstyle N=\max _{j}N_{j}}...

\forall n>N_{1}.\forall \varepsilon >0\exists N_{2}(\varepsilon ):|a_{n}-b|<\varepsilon ,\forall n>N_{2}.} ε = b − a 2 > 0 N = max { N 1 ( b − a 2 ) , N 2 (...

{\displaystyle \exists } N {\displaystyle N} = {\displaystyle =} m a x {\displaystyle max} ( {\displaystyle (} N 1 {\displaystyle N_{1}} ( ε ) {\displaystyle...

{ Im ⁡ z > 0 , | z | > r } ) ; max z ∈ γ R | f ( z ) | → 0 {\textstyle f\in C(\{\operatorname {Im} z>0,|z|>r\});\max _{z\in \gamma _{R}}|f(z)|\rightarrow...

fertility=1.0) children = concatenate((children, illegals)) return children, max_children N = 1000000 male_portion = 0.51 fertility = 0.92 law_breakers = 0.06 wanted_children...

если max 0 ≤ n ≤ N | y n ∗ − y n | ≤ k ( τ ) max 0 ≤ i ≤ k − 1 | ε i | {\displaystyle \max _{0\leq n\leq N}|y_{n}^{*}-y_{n}|\leq k(\tau )\max _{0\leq...

Mülheim an der Ruhr […] n Мюльхайм-на-Руре Примеры: In Mülheim an der Ruhr befindet sich das Max-Planck-Institut für bioanorganische Chemie....

∃ a ∈ D | f ( a ) | = max z ∈ D | f ( z ) | ⇒ f ( z ) ≡ c o n s t {\textstyle f\in O(D).\quad \exists a\in D\quad |f(a)|=\max _{z\in D}|f(z)|\Rightarrow...

_{max}}})} . Оптимально τ = 2 λ m i n + λ m a x {\displaystyle \tau ={\frac {2}{\lambda _{min}+\lambda _{max}}}} Тогда ‖ B ‖ 2 = λ m a x − λ m i n λ m...

{\displaystyle \exists L=max_{(x,y)\in S}|f(x,y)|<\infty } . Зафиксируем некоторое N и рассмотрим разбиение отрезка: x i = x 0 + i ∗ h n , h n = △ N {\displaystyle...

\gamma =\max _{x,y\in \gamma }\rho (x,y)\end{aligned}}} Оценим знаменатель ( ξ − z − h ) n + 1 ( ξ − z ) n + 2 {\textstyle (\xi -z-h)^{n+1}(\xi -z)^{n+2}}...

max y ∈ ∂ Ω | U 1 ( x ) U 2 ( x ) | < ε {\displaystyle f_{2}:\max {y\in \partial \Omega }|U_{1}(x)U_{2}(x)|<\varepsilon } , из следствия 2: f 2 : max...

− 1 | , | x n − ε | , | x n + ε | } {\displaystyle A=max\{{|x_{1}|,|x_{2}|,|x_{3}|,...,|x_{N-1}|,|x_{n}-\varepsilon |,|x_{n}+\varepsilon |\}}} . В отрезке...

n ) = ( g , e n ) + ( h , e n ) ⏟ 0 , 1 ≤ n ≤ N ⇒ ( f , e n ) = ( g , e n ) = ( ∑ n − 1 N c n e n , e n ) ⇒ c n = ( f , e n ) {\displaystyle (f,e_{n})=(g...

{\frac {2\pi r\max _{|\xi -z|\pm r}|f(\xi )-f(z)|}{r}}=} = 2 π max | ξ − z | = r | f ( ξ ) − f ( z ) | → 0 ,  r → 0 {\textstyle =2\pi \max _{|\xi -z|=r}|f(\xi...

max 1 ≤ j ≤ m ∑ i = 1 m | a i j | , ‖ A ‖ 2 = max 1 ≤ j ≤ m λ j ( A T A ) , ‖ A ‖ ∞ = max 1 ≤ j ≤ m ∑ i = 1 m | a j i | {\displaystyle \|A\|_{1}=\max...

обусловленности (относительное). max x ≠ 0 ν δ ( x ) = ‖ A − 1 ‖ ⋅ ‖ A ‖ = ν ( A ) = c o n d ( A ) {\displaystyle \max {x\neq 0}\nu _{\delta }(x)=\|A^{-1}\|\cdot...