lecture_25 - MA 36600 LECTURE NOTES MONDAY MARCH 23 Higher...

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MA 36600 LECTURE NOTES: MONDAY, MARCH 23 Higher Order Linear Equations Linear Independence. Consider a collection of functions { f 1 , f 2 , . . . , f n } . We say that this is a linearly independent set if the only solution to the equation k 1 f 1 ( t ) + k 2 f 2 ( t ) + · · · + k n f n ( t ) = 0 for all t is k 1 = k 2 = · · · = k n = 0. We say that this is a linearly dependent set otherwise. We explain the relationship with the Wronskian , as the n × n determinant, W ( f 1 , f 2 , . . . , f n ) ( t ) = f 1 ( t ) f 2 ( t ) · · · f n ( t ) f (1) 1 ( t ) f (1) 2 ( t ) · · · f (1) n ( t ) . . . . . . . . . . . . f ( n - 1) 1 ( t ) f ( n - 1) 2 ( t ) · · · f ( n - 1) n ( t ) . We will show that the following are equivalent: i. W ( f 1 , f 2 , . . . , f n ) ( t 0 ) 6 = 0 for some t 0 . ii. W ( f 1 , f 2 , . . . , f n ) ( t ) 6 = 0 for all t . iii. { f 1 , f 2 , . . . , f n } is a linearly independent set. We give the proof, following Lecture #16 on Monday, February 23. Consider the following propositions: p 1 = “ W ( f 1 , f 2 , . . . , f n ) ( t 0 ) 6 = 0 for some t 0 p 2 = “ W ( f 1 , f 2 , . . . , f n ) ( t ) 6 = 0 for all t p 3 = “ { f 1 , f 2 , . . . , f n } is a linearly independent set” We have already seen that p 1 ⇐⇒ p 2 ; this is Abel’s Theorem: Say that { y 1 , y 2 , . . . , y n } is a set of functions satisfying the equation P 0 ( t ) d n y dt n + P 1 ( t ) d n - 1 y dt n - 1 + · · · + P n - 1 ( t ) dy dt + P n ( t ) y = 0 . Then their Wronskian satisfies P 0 ( t ) dW dt + P 1 ( t ) W = 0 = W ( y 1 , y 2 , . . . , y n ) ( t ) = C exp - Z t P 1 ( τ ) P 0 ( τ ) .
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