Huge linear system of equations with powers of 2: (2^0)^n a_n+(2^0)^(n-1) a_(n-1)+...+(2^0)^1 a_1=4^0, (2^1)^n a_n+(2^1)^(n-1) a_(n-1)+...+(2^1)^1 a_1=4^1 ... (2^(n-1))^(n-1) a_n+(2^0)^(n-1) a_(n-1)+...+(2^^(n-1))^1 a_1=4^^(n-1)

Buszmenan 2022-09-01 Answered
Huge linear system of equations with powers of 2:
( 2 0 ) n a n + ( 2 0 ) n 1 a n 1 + + ( 2 0 ) 1 a 1 = 4 0 ( 2 1 ) n a n + ( 2 1 ) n 1 a n 1 + + ( 2 1 ) 1 a 1 = 4 1 ( 2 n 1 ) n a n + ( 2 n 1 ) n 1 a n 1 + + ( 2 n 1 ) 1 a 1 = 4 n 1
for n 2. Show that the (unique) solution of this system is when a 2 = 1 and all other variables are zero.
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Answers (2)

ordonansexa
Answered 2022-09-02 Author has 7 answers
Suppose that ( a 1 , , a n ) is a solution of the proposed system. Consider P ( X ) = X 2 + 1 n a j X j . This is a polynomial of degree smaller or equal to n, with P ( 0 ) = 0. Moreover, by assumption P ( 2 i ) = 0 for every i = 0 , n 1, so P has at n + 1 zeros, and it must be identically zero. So ( a 1 , a 2 , , a n ) = ( 0 , 1 , , 0 ). The converse is trivially true.
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hikstac0
Answered 2022-09-03 Author has 2 answers
Use Cramer's rule remembering the 4 k = ( 2 k ) 2 which makes the right hand side the same as the second to last column. This means that two columns will be the same (and so the determinant will be zero) for all variables except for a 2 where the determinant in the numerator will be the same as the determinant in the denominator.
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