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131A Exam1 Sol

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  MIDTERM EXAM IProblem 1.  Let   b  and   c  be two real numbers and consider the function   f  on   R  defined by  f  ( x ) =  x 2 + 2 bx  +  c, x  ∈  R . Let   A  be the set defined by  A  =  { x  ∈ R :  f  ( x )  ≥  0 } . As the set   A  depends on the parameters   b,c  ∈  R , we write   A ( b,c )  for   A  toindicate the dependence of the set   A  on   ( b,c )  ∈ R 2 . Answer to the following problems  :1-a)  Show that   A ( b,c )   =  ∅  for every pair   ( b,c )  ∈ R 2 .Solution.  The graph of the function  f  ( x ) =  x 2 + 2 bx  +  c  is an upward openparabola. The question is asking if the function  f   takes non-negative valueat some point in  R . Of course, it takes positive values at a certain value of  x  regardless of what values of   b  and  c  are assigned for  f  . For example if  c  ≥  0, then  f  (0) =  c  ≥  0, so that 0  ∈  A ( b,c ), and if   c <  0 then  b 2 − c >  0,so that √  b 2 − c  is a real number and consequently  f  ( − b + √  b 2 − c ) = 0 and − b  + √  b 2 − c  ∈  A ( b,c ). Therefore in either case,  A ( b,c ) is non-empty.  ♥ 1  2  MIDTERM EXAM I 1-b)  Let   B  be the set of pairs   ( b,c )  ∈  R 2 such that   A ( b,c ) =  R .  Draw the set   B  on the   bc −  plane : Solution.  First we observe that A ( b,c ) = R means exactly  f  ( x )  ≥  0 forevery  x  ∈  R . Thus the set  B of ( b,c ) such that  A ( b,c ) =  R is the exactly the set of pairs( b,c ) such that x 2 +2 bx  +  c  ≥  0  bc for every  x  ∈ R . As the function  f   can be written in the form: f  ( x ) = ( x  +  b ) 2 +  c − b 2 , x  ∈  R , for  f  ( x ) to be non-negative everywhere it is necessary and sufficient that  c − b 2 ≥  0. Thus thegraph of the set  B  looks like the above shaded area with  c -axis virtical and  b -axis horizontal.The boundary curve  c  =  b 2 is included in the set  B .  ♥ Problem 2.  Let   { a n }  and   { b n }  be two convergent sequences of real numbers such that  lim n →∞ a n  =  a  = lim n →∞ b n . Define a new sequence   { c n }  by the following  :   c 2 n − 1  =  a n , n  = 1 , 2 , ···  ; c 2 n  =  b n , n  = 1 , 2 , ···  . Prove that the new sequence   { c n }  converges to  a .Proof.  From the convergence: lim a n  =  a , it follows that for any  ε >  0 there exists  N  a ( ε )  ∈ N such that | a n  − a |  < ε  for every  n  ≥  N  a ( ε ) .  MIDTERM EXAM I  3 Similarly for any  ε >  0 there exists  N  b ( ε )  ∈ N  such that | b n  − b |  < ε  for every  n  ≥  N  b ( ε ) . Suppose  n  ≥  2max { N  α ( ε ) ,N  b ( ε ) } . Then we have  n/ 2  ≥  N  b  for even  n  and ( n +1) / 2  ≥  N  a for odd  n  and also c n  =   a ( n +1) / 2  if   n  is odd; b n/ 2  if   n  is even . Hence we conclude that | c n  − a |  < ε  if   n  ≥  2max { N  α ( ε ) ,N  b ( ε ) } . Thus  N  c ( ε ) = 2max { N  α ( ε ) ,N  b ( ε ) }  serves as the waiting time for  { c n }  to come closer to  a within  ε  distance.  ♥ Problem 3.  Prove the identity  1 3 +2 3 + ··· + n 3 = (1 +2 + 3 + ··· +  n ) 2 , n  ∈ N , by mathematical induction.Proof.  For  n  = 1, the both sides of the formula equal to 1. So it holds definiety for  n  = 1.Suppose that1 3 +2 3 + ··· + n 3 = (1 + 2 + 3 + ··· +  n ) 2 . We add ( n  + 1) 3 to the left side and obtain  1 3 + 2 3 + ··· +  n 3   +( n  + 1) 3 = (1 + 2 + 3 + ··· +  n ) 2 + ( n  + 1) 3 by the induction assumption=  n ( n  + 1)2  2 + ( n  + 1) 3 by Proposition 1.4.3=  n 2 ( n  + 1) 2 4   + ( n  + 1) 2 ( n  + 1)= ( n  + 1) 2  n 2 4 + ( n  +1)   = ( n  + 1) 2 n 2 + 4 n  + 44= ( n  + 1) 2 ( n  + 2) 2 4 =  ( n  + 1)( n  + 2)2  2 = (1 + 2 + 3 + ··· +  n  + ( n  + 1)) 2 by Proposition 1.4.3 . The last term is exactly the right hand side of the formula for  n + 1. Therefore the mathe-matical induction guarantees the validity of the formula for all natural number  n .  ♥
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