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Discrete Math. Practice. For students of technical specialties
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Solution: Let m = | √ |x||
Because the rules ≤ |x|=n <=> nx <n1 should m ≤ √ |x| <m+1
m
≤ |x| (<m+1)
From the rules: x <n <=> |x| <n
n ≤x <=> n <|x|,where nan integer
must be m
≤ x (<m+1)
m ≤ √ x <m+1
From (*) => m= | √
Thus | √ |x||= m= | √x|.
Problem No. 31: u
= -u
+2u
u
= 0, u
= 1
Solution: here K (x) =1+x-2x
. We calculate: D (x) =K (x) u (x) = (1+x-2x
) (u
+ u
x+…) 0+x+0=x
We get: u (x) = ((x) \ (1+x-2x
))
The next step is the expansion of the denominator K (x) to the product (a
x) (1a
x), then 〈
and 〈
– the roots of the quadratic equation: 1+x-2x
=0
and
a
=1
a
= ((-1) \ (2))
arrive to the formula: u (x) = ((x) \ (12+x-x
)) = ((x) \ ((1-x) (1+ ((x) \ (2)))))
We find the decomposition into a sum of simple fractions by the method of indefinite coefficients: ((x) \ ((1-x) (1+ ((x) \ (2))))) = ((A) \ ((1-x))) + ((B) \ (1+ ((x) \ (2))))
We obtain a system of linear equations: ((1) \ (2)) AB=1B+A=0
Its solution: A= ((2) \ (3)),B= ((-2) \ (3)). Hence: u (x) = (((2) \ (3)) \ (1-x)) + (((-2) \ (3)) \ (1- ((x) \ (2)))) = ((2) \ (3)) ∑
1
x
– ((2) \ (3)) ∑
((-1) \ (3))
x
This leads to the answer: u
= ((2) \ (3)) – ((2) \ (3)) * ((-1) \ (2))
Answer: u
= ((2) \ (3)) – ((2) \ (3)) * ((-1) \ (2))
Task number 32: 15 students shook hands at a meeting of students, three people made 4 handshakes, and others – 3. How many students were there.
Последний отзыв
Книга дельная. Без воды, приводятся реальные кейсы и способы их разрешения. Много практичных полезных советов и методик. Кое-что уже испробовала в дел...
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