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Texas A&M University
Mathematics
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MATH 152 Exam III - - Spring 1996

Part I - - Multiple Choice

No calculators. No partial credit.


1. The series 

                            infinity
                             -----       n
                              \      (-1)
                               )     -----
                              /       1/4
                             -----   n
                             n = 1

is

a) harmonic   b) divergent  c) absolutely convergent  d) geometric  e) conditionally convergent.

SOLUTION:  e)   The terms 1/n^(1/4) monotonically converge to zero; so
this series converges by the alternating series test. This series does not 
converge absolutely because the exponent 1/4 is less than or equal to one
(the integral test implies that the series involving 1/n^p converges for p>1).


2. The curve parametrized by  x= e^t  y=e^{-t}  for 0<t<1,  may be written as:

a) y = e^x  for  0<x<1    b) y = 1/x  for   1<x<e    c) y= -x  for  1<x<e
              d)  y = e^{-x)$  for  1<x<e    e) y = 1/x,  for  0<x<1

SOLUTION:  b)  Note that y = e^(-t) =  1/e^t = 1/x  and that when t=0, x = 1
and when t = 1, x= e.


3. The function  f(x)  defined by the series

                               infinity
                                -----    (2 n)
                                 \      x
                       f(x) :=    )     ------
                                 /      (2 n)!
                                -----
                                n = 0


satisfies which of the following differential equations?

a)  f'' = - 2 f     b)  f'' = - f     c)  f'' = 0   d)   f'' = f    e)  f'' = 2 f

SOLUTION:  d)   The second derivative of  x^(2n)/(2n)!  is
 (2n)(2n-1)x^(2n-2) / (2n)!. = x^(2n-2)/(2n-2)!    Since (2n)! = (2n)(2n-1)(2n-2)!
                                           = x^(2m)/(2m)!         by letting  m = n-1
So the second derivative of  f  equals 

                           infinity
                            -----    (2 m)
                             \      x
                              )     ------
                             /      (2 m)!
                            -----
                            m = 0

which is another way of writing the function  f.




5. The series 

                 infinity
                  -----
                   \
                    )     (arctan(n + 1) - arctan(n))
                   /
                  -----
                  n = 1

equals

a) Pi/8    b) Pi/4   c) Pi/2   d) Pi  e) does not converge


SOLUTION:   b)   This series telescopes to  arctan(infinity) - arctan(1)
(or more precisely to  the limit of   arctan(N) - arctan(1)  as N -> infinity).
Since arctan(infinity) = Pi/2   and  arctan(1) = Pi/4,  the answer is 
Pi/2 - Pi/4 = Pi/4.


6. Find the radius of convergence of the series 
 
> Sum((x-4)^n/n^n,n=1..infinity);

                          infinity
                           -----          n
                            \      (x - 4)
                             )     --------
                            /          n
                           -----      n
                           n = 1


a)  0      b)  1/2      c) 1      d)  2      e) infinity

SOLUTION:  e)   Use  the root test - - the nth root of the nth term
is  |x-4|/n  which converges to zero for any value of  x.  Therefore the 
root test implies that  this series converges for any value of  x.  In other
words, the radius of convergence is infinity.


7. Find the coefficient of  (x-2)^2  in the Taylor series expansion of 
g(x) = 1/x  about  x=2.

a) 1/2   b) 1/4   c)  1/8   d) 1/16   e)  1/32

SOLUTION:  c)  The coefficient is  g''(2)/2;  since g''(x) = 2/x^3 inserting
x = 2  yields  g''(2) = 2/8 = 1/4.  Therefore  g''(2)/2 = 1/8.


8. The line tangent to the curve  x = t^3-t  y = t^2  at the point  (6,4)  has slope

a) 8/107    b) 4/11    c) 2/3    d) 3/2     e) 47

SOLUTION:  b)  The point (6,4) corresponds to t = 2.  The slope of the tangent
at (6,4) is therefore equal to y'(2)/x'(2); since  x' = 2t  and y' = 3t^2 - 1,
the slope is 4/11.


9. Find the coefficient of   x^3  in the Maclaurin series expansion 
(Taylor series about x=0) of   
 
> g(x) := (1-x^2/2)*exp(x);

                                       2
                     g(x) := (1 - 1/2 x ) exp(x)


a) -1/3    b) 0     c) 1/2     d) 1     e) 3/2

SOLUTION:    a)   Expanding  e^x, we get
    g(x) = (1 - x^2/2) (1 + x + x^2/2! + x^3/3! + ...)
The  x^3  term  of this product equals  x^3/3! - x^3/2 = -x^3/3.
Therefore the coefficient of x^3 is  -1/3.

10. On which of the following intervals does the series 
 
                          infinity
                           -----
                            \             n
                             )     (x - 4)
                            /
                           -----
                           n = 1

converge?

a)  (-1,1)    b) (0,8)    c) (3,5)  d)  (-1/2, 1/2)    e) (-infinity, infinity)

SOLUTION:  c)  This is a geometric series, which converges provided
|x-4| < 1  or another words,  3 < x < 5.

11. Suppose that 

                                infinity
                                 -----
                                  \
                                   )     a_n
                                  /
                                 -----
                                 n = 1


                                infinity
                                 -----
                                  \
                                   )     b_n
                                  /
                                 -----
                                 n = 1


are series with positive terms such that  a_n/b_n  --> 0 ans n - -> infinity. 
 Which statement is true?

a)  If  


                             infinity
                              -----
                               \
      a)                If      )     b_n        converges then
                               /
                              -----
                              n = 1

                             infinity
                              -----
                               \
                                )     a_n     converges
                               /
                              -----
                              n = 1


                             infinity
                              -----
                               \
       b)            If         )     b_n  diverges, then
                               /
                              -----
                              n = 1

                             infinity
                              -----
                               \
                                )     a_n  diverges
                               /
                              -----
                              n = 1


                             infinity
                              -----
                               \
       c)            If         )     a_n   converges, then
                               /
                              -----
                              n = 1


                             infinity
                              -----
                               \
                                )     b_n    diverges
                               /
                              -----
                              n = 1


                             infinity
                              -----
                               \
        d)          If          )     b_n   converges, then
                               /
                              -----
                              n = 1


                             infinity
                              -----
                               \
                                )     a_n    diverges.
                               /
                              -----
                              n = 1


                             infinity
                              -----
         e)         If         \
                                )     a_n   converges, then
                               /
                              -----
                              n = 1


                             infinity
                              -----
                               \
                                )     b_n    converges.
                               /
                              -----
                              n = 1


SOLUTION:  a)   If  a_n/b_n -->  0,  then eventually, the a_n
are all  smaller than the b_n.  This means that if the series involving
b_n converges,  then the series involving a_n  must also converge.

Part II - - Work Out Problems

Calculators are permitted. Partial credit may be given.


12.  Find the Maclaurin series for  1/(1+x^5).  Use this series
to approximate


                              .1
                             /
                            |      66
                            |    ------ dx
                            |         5
                           /     1 + x
                             0

 
SOLUTION:   The Maclaurin series is



                        infinity
                         -----
                          \          n  (5 n)
                           )     (-1)  x
                          /
                         -----
                         n = 0

using the formula for a geometric series.
  To approximate the integral, the idea is to break up this series into
a sum of the first N terms and a remainder which is a sum from N+1 to infinity.
Then we integrate the first N terms (which is a polynomial). The trick is to find
a convenient value of  N  so that the integral of the remainder is less than 10^(-14).
Now the remainder is a geometric series whose value is 


                                (5 N + 5)
                               x
                            66 ----------
                                      5
                                 1 + x


which is less than  66 x^(5N+5). Integrating this over 0 to 0.1, gives
66 10^(-5N-6)/(5N+6). By trial and error, we see that N=2 gives
(66/16)(10)^(-16) which is less than 10^(-14). Therefore, the approximate
value for the integral is obtained by computing


                       1/10
                      /
                     |               5       10
                     |      66 - 66 x  + 66 x   dx
                     |
                    /
                      0


which is 6.59998900006.


13. Find the radius of convergence of the series 
 

                           infinity
                            -----    (2 n)
                             \      x
                              )     ------
                             /         n
                            -----     2
                            n = 1


SOLUTION:   Write the series as

                          infinity
                           -----
                            \            2 n
                             )     (1/2 x )
                            /
                           -----
                           n = 1

This is a geometric series which converges when x^2/2 < 1,
or  x^2 < 2 which is equivalent to -sqrt(2) < x < sqrt(2).
So the radius of convergence is sqrt(2).


14. Determine whether the series


                         infinity
                          -----
                           \          1
                            )     ----------
                           /       2       n
                          -----   n  + (-1)
                          n = 2

converges.  Justify your answer.

SOLUTION:   This series converges by comparison with the series

                            infinity
                             -----
                              \       1
                               )     ----
                              /        2
                             -----    n
                             n = 2

which converges by the integral test.


15. What is the smallest value of  n  for which we can be sure that 
the  nth   partial sum 


                               n
                             -----
                              \    / 5  \
                               )   |----|
                              /    |  6 |
                             ----- \ k  /
                             k = 1

approximates

                           infinity
                            -----
                             \      / 5  \
                              )     |----|
                             /      |  6 |
                            -----   \ k  /
                            k = 1

 to within  10^(-10)?
HINT: Use the integral test to bound the remainder.

SOLUTION:   By the integral test, the remainder

                           infinity
                             -----
                              \      / 5  \
                               )     |----|
                              /      |  6 |
                             -----   \ k  /
                           k = n + 1

is less than


                            infinity
                           /
                          |           5
                          |          ---- dx
                          |            6
                         /            x
                           n

which is


                                  1
                                 ----
                                   5
                                  n

This remainder is within 10^(-10) provided  n  is at least 10^2 = 100.

16. The point  A=(a,0)  is moving on the positive  x - axis 
and the point    B = (0,b)  is moving on the positive y - axis so 
that the distance between  A  and  B  is always 5. Let  C  be the 
midpoint of  the line segment from  A  to  B.. Find parametric equations 
for the curve traced out by C, using b as the parameter.

SOLUTION:   The midpoint  from  A to B is  x = a/2 ,   y = b/2.
We must describe these equations in terms of  the parameter  b.
From the given information,  a^2+b^2 = 25  (Pythagorean Theorem). So

                                      2 1/2
                           b = (25 - a )

and so the point is


                                        2 1/2
                         x = 1/2 (25 - a )


                              y = 1/2 b


17. What is the length of the curve parametrized by 
x = t^2/2 - t    y = (4/3) t^(3/2)
between  t = 0  and   t  =  2?

SOLUTION:    The arclength is 



                    2
                   /
                  |   //d      \2   /d      \2\1/2
                  |   ||-- x(t)|  + |-- y(t)| |    dt
                  |   \\dt     /    \dt     / /
                 /
                   0

Inserting  x = t^2/2 - t    y = (4/3) t^(3/2),  the integrand  becomes
sqrt( (t-1)^2 +4t)  =  sqrt(t^2+2t+1)  =  t+1.  Therefore the length is



                               2
                              /
                             |
                             |   1 + t dt
                             |
                            /
                              0


which is 4.