goldch6 - Contents VI Asymptotic Behaviour of Rational...

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Contents VI Asymptotic Behaviour of Rational Functions 1 VI.A Rational Functions as x → ±∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 VI.B Rational Functions Near Zeros of the Denominator; Poles . . . . . . . . . . . . . . . . . . . 7 VI.C Graph Versus Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
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VI Asymptotic Behaviour of Rational Functions VI.1 VI Asymptotic Behaviour of Rational Functions VI.A Rational Functions as x → ±∞ There exists a positive real constant R such that all roots of Q ( x ) are in [ - R, R ] and so f ( x ) = P ( x ) Q ( x ) is defined whenever x < - R or R < x , in other words, x is outside the interval [ - R, R ]. We’ll look at the behaviour of f ( x ) when i) x < - R and x → -∞ , or ii) R < x and x → ∞ . Property 1 : If deg P < deg Q then f ( x ) 0 as x → ±∞ . Proof: Write P ( x ) = n X i =0 a i x i ; a n 6 = 0 and Q ( x ) = m X j =0 b j x j ; b m 6 = 0 . We have assumed m > n . Thus dividing by x m we find for x 6 = 0 f ( x ) = (1 /x m ) P ( x ) (1 /x m ) Q ( x ) = n i =0 a i x i - m m j =0 b j x j - m . The numerator : This has the form a n x n + a n - 1 x n - 1 + · · · + a 0 x m = a n x m - n + a n - 1 x m - n +1 + · · · + a 0 x m . Since m > n each m - i > 0. Thus, as | x | → + , | x m - i | → + and a i x m - i 0. Therefore the numerator goes to 0 as x → ±∞ . The denominator : This has the form b m x m + · · · + b 0 x m = b m + b m - 1 x + · · · + b 0 x m b m as x → ±∞ . Therefore, the denominator goes to b m as x → ±∞ . The limit of f : f ( x ) = numerator denominator 0 b m = 0 as x → ±∞ because b m 6 = 0. Example . For f ( x ) = x 2 +10 10 x 3 + x +1 , deg(numerator) - deg(denominator) = 2 - 3 = - 1 < 0. Therefore f ( x ) 0 as x → ±∞ . Indeed,
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VI.A Rational Functions as x → ±∞ VI.2 f ( x ) = x 2 +10 10 x 3 + x +1 = 1 x 3 ( x 2 +10 10 ) 1 x 3 ( x 3 + x +1) = 1 x + 10 10 x 3 1+ 1 x 2 + 1 x 3 0+0 1+0+0 = 0 as x → ±∞ . Property 2 : If P ( x ) = n i =0 a i x i , a n 6 = 0 and Q ( x ) = n j =0 b j x j , b n 6 = 0, have same degree then f ( x ) a n b n 6 = 0 as x → ±∞ . Proof: Using long division, we divide Q ( x ) into P ( x ), to get P ( x ) = a n b n Q ( x ) + R ( x ) , with deg R < n , or R = 0.
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