must be m2 x (<m+1) 2
m x <m+1
From (*) => m= |
Thus | |x||= m= | x|.
Problem No. 31: un +2 = -un +1 +2un, u0 = 0, u1 = 1
Solution: here K (x) =1+x-2x2. We calculate: D (x) =K (x) u (x) = (1+x-2x2) (u0 + u1x+) 0+x+0=x
We get: u (x) = ((x) \ (1+x-2x2))
The next step is the expansion of the denominator K (x) to the product (a11x) (1a2x), then 1 and 2 the roots of the quadratic equation: 1+x-2x2=0
and
a1=1
a1= ((-1) \ (2))
arrive to the formula: u (x) = ((x) \ (12+x-x2)) = ((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)) k=01kxk ((2) \ (3)) k=0 ((-1) \ (3)) kxk
This leads to the answer: un= ((2) \ (3)) ((2) \ (3)) * ((-1) \ (2)) n
Answer: un= ((2) \ (3)) ((2) \ (3)) * ((-1) \ (2)) n
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.
Solution: We will consider a case requiring a smaller number of participants. Since the three made 4 handshakes: we consider. That between them they made maximum handshakes two, and formed a complete 3x vertex graph and used only 3 handshakes out of the total:
Each «had» two handshakes. The minimum number of vertices that must be added so that the condition «three made 4 handshakes» is met these are two. Lets portray them like this:
We get already 9 used handshakes. And the two added vertices have 3 handshakes, which corresponds to the condition. It remains 6. Just the full 4-vertex graph gives us 6 handshakes for each of the 4 students added: