MCS 471 Extra Practice Problems for Nonlinear Equations with Some Answers
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CAUTION: Some of these old exam problems and answers may not be relevant to
         your current course.   Caveat Usor!

   In the following problems, use "EXAM PRECISION:  Chop to 4 significant
(4C) digits only when you write an intermediate or final answer down and
continue calculations with those numbers recorded.  These problems are
just for practice and are not to be handed in.
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1.  Using the method of BISECTION starting with A(1)=1. and B(1)=2., find
the root of
                             F(X)=EXP(X)-3/X.
Put your answer in a table of
                        K,A(K),B(K),F(A(K)),F(B(K))
for each iteration on (A(K),B(K)) for K=1 to 3.
(Final Ans.:  Root =(4C) (1,1.125) )
 
2.  Find the root of
                              F(X)=EXP(X)-3/X
using the SECANT METHOD for 2 iterations beyond the starting guesses,
X(1)=1. and X(2)=2.  Record your answers in a table of
                      K,X(K),X(K-1),F(X(K)),F(X(K-1))
for each iteration  K.
(Final Ans.:  Root =(4C) 1.049 )
 
3.  Find the root of
                         F(X)=EXP(X)-3/X on (1,2)
using NEWTON'S METHOD until ABS(X(K)-X(K-1))<.5E-1.  Record Results in
table of
                              K,X(K),F(X(K)).
Use X(1)=1.5 to start.
(Final Ans.:  root=(4C) 1.049)
 
4.  Find the relative error with chopping to 4 significant digits,
recording the Results of each binary operation and using full calculator
precision as a substitute for the exact value:
                (.3876E-13*.4823E+11-.7526E+08*.2552E-10).
(Final Ans.:  .4691E-00%)
 
5.  Numerically solve
                             F(X)=LN(X)-1/X=0
by forming a convergent, fixed point iteration, other than Newton's,
starting from  X(1)=EXP(1).  Record your answers in a table of
                         K, X(K),  for K= 1 to 3.
(Final Ans.=(4CH)  (3,1.998))
 
6(ME/A/W83).  Approximate the root of
                            F(X)=EXP(X)-5/X**2
using the SECANT METHOD and starting iterates: X(1)=1 and X(2)=2.
However, modify the usual algorithm for this method so that each new
iterate is calculated from the two prior iterates with the smallest
absolute values of the given function.  For each iterate X(K+1) record
    K, X(K-1), X(K), F(X(K-1)), F(X(K)),ABS(X(K)-X(K-1))+ABS(F(X(K))).
Stop iterating when the last item in your recorded table is less than
0.1E-1.
(Final Ans.:  (6,1.216,1.216,-.7783E-2,-.7783E-2,+.7783E-2))
 
7(ME/A/F82). (A) Establish the quadratic convergence property of NEWTON'S
METHOD for finding zeros of functions, F(Z)=0, i.e., show that the
(K+1)st error E(K+1)=X(K+1)-Z=C*E(K)**2 (approx.) for X(K) near Z BY
finding C in terms of the derivatives of F at Z.  Assume F'(Z) NOT= 0
and F"(Z) exists.
(B) Apply NEWTON'S METHOD to find the positive zero of
                           F(X)=2*COS(X)-EXP(X)
for some K so that
                  ABS(X(K)-X(K-1))+ABS(F(X(K-1)))<0.5E-3.
for each K.  Record
                            K, X, F(X), F'(X).
(Final Ans.:  (6,.5397,.2336E-3,-2.743))
 
8.   Find the minimum of  G(X)=EXP(X)+8/X  on (1,2) by the method of
GOLDEN SECTION SEARCH for K=1,2,3 iterations. Display your answer in a
table of
                       K, AK, BK, XK, UK, GXK, GUX.
(Final Ans.:  MIN(G)=(4C) 9.771  at X(MIN)=(4C) 1.382 )
 
9.  Find the maximum and its location for G(X)=X*COS(X) on (0.4,1.4) by
the method of GOLDEN SECTION SEARCH.  Summarize your results by a
tabulation of
                        K, AK, BK, XK, UK, GXK, GUK
for K=1 to 3.
(Best Final Ans.:  (.8720,.5609)  for (X,G) OR (U,G). )
 
10.  Using the method of GOLDEN SECTION SEARCH find maximum and
location of
                      G(X)=ALOG(X)/(1+X**2)  on (1,3).
Summarize your Results by tabulation of
                        K, AK, BK, XK, UK, GXK, GUK
for K= 1 to KMAX, where ABS(BK-AK)<.3 at K=KMAX.  At KMAX, state your
best current maximum and its location and the interval of uncertainty.
  (Ans.:
      K    AK     BK     XK     UK     GXK    GUK  BK-AK
      1  1.000  3.000  1.764  2.236  .1380  .1341  2.000
      2  1.000  2.236  1.472  1.764  .1220  .1380  1.236
      3  1.472  2.236  1.764  1.944  .1380  .1390  .764
      4  1.764  2.236  1.944  2.055  .1390  .1379  .472
      5  1.764  2.055  1.875  1.944  .1392  .1390  .291<.3
 KMAX=5;   MAX(G)=(4C) .1393;  XMAX=(4C) 1.875;  ABS(BMAX-AMAX)=.291
 or alternately MAX(G)=(4C) .1392; XMAX=(4C) 1.874; Int. Uncern.=0.292)
 
11.  Using Newton's Method, find a numerical approximation to
intersection
                           EXP(x/2) = LN(x) + 3
on [2,3] starting with the endpoint having the smallest value of |F|,
tabulating
            {K, X(K), F(X(K)), F'(X(K)), X(K+1), |X(K+1)-X(K)|}
for K = 1 to 2.
 
12.  Find a numerical approximation to the intersection of
                              LN(x) = EXP(-x)
starting from X(1) = 0 at K = 1 using a convergent FIXED POINT ITERATION
other than Newton's method.  Prove that the iteration is convergent.
Calculate a few iterates and tabulate {K,X(K)} for K = 1 to 3.

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