MCS 471 Practice Problems 1: Nonlinear Equations

Background Reading:

• Lecture Notes, Class Notes

• Heath, Chapters 1 and 5

• Gerald and Wheatley, Chapters 0 and 1

• Class Maple Web Home Page, introductions and fsolve

  • Beginner's Introduction to MapleV Release 5
  • MSCS Maple Help Page (Prof. Hurder)
  • MSCS Maple Equations (fsolve) Lab
  • MSCS Maple Plot Lab
  • MCS 471 Help for Maple Symbolic Tools

• Class MATLAB Web Home Page, introductions and fzero
  • Hanson's MATLAB Help Page

• Exam 1 Topics

1. Floating Point Arithmetic (for advances information, see Floating Point Arithmetic References

2. Bisection Method

3. Secant Method

4. Newton's Method

5. Fixed Point Iteration

6. Golden Section Search, (See Lecture Notes and or click for Class Notes, but topic is briefly in Heath text)

Practice Problems:

1. Considering just the field representing the fractional part of a nonzero normalized 32 bit floating point real for the IEEE 754 single precision standard, how many distinct numbers can be represented assuming the exponent and sign are fixed? Record as many digits as your calculator allows. Explain how you got your answer. (I.e., ignore the sign bit and the biased exponent field for this count.)

   ---

   Answer: Since

   fl[x] = ±(1+frac) * 2^(exp)

   for normalized floating point numbers with the leading "1" not being stored and the "frac" integer is stored as a 23 bit integer

   0   <   frac_2,32,q   <   2^(23)-1,

   
   
   the latter bound being a string of 23 ones, so that the any digit can be a binary 0 or 1 and the number of numbers stored is

   2^(23) = 8,388,608

   
   
   for the 32 bits answer.

2. Using Newton's Method, find a numerical approximation to the Zero of

   f(x) = e^(-x/3) - 0.05471*x

   
   
   on [2.5,2.9] starting with the endpoint (k = 0) having the smallest value of |f|, keeping track of the number of ALL function evaluations kfe, the current change in sign interval [a_k,b_k], and tabulating

   k	kfe	a_k	b_k	x_k	f_k	f_k'	x_{k+1}	|x_{k+1}-x_k|
   0
   ...								...

   until |x_{k+1}-x_k| < 0.5e-2.

   ---

   Answer: Examination of the function f(x) indicates that there is no change of sign in [2.5,2.9] and since Newton's method does not depend on it, we proceed ignoring the change in sign interval:

   f'(x)=-e^(-x/3)/3- 0.05471 ,   x_{k+1} = x_k   -   f_k/f_k'; x_0=b=2.9=argmin[f(a),f(b)]=(r4) argmin[0.2978,0.2217]; note that kfe=3 on the first step since f(a), f(b) and f'(b) are 3 function evaluations.

   k	kfe	a_k	b_k	x_k	f_k	f_k'	x_{k+1}	|x_{k+1}-x_k|
   0	3	2.5*	2.9*	2.9	0.2217	-0.1815	4.121	1.221
   1	5	4.121?	?	4.121	2.772e-2	-0.1391	4.320	0.1990
   2	7	4.320?	?	4.320	5.806e-4	-0.1337	4.324	0.4000e-2

   Stopping since tol=0.5e-2 is satisfied.

3. Find a numerical approximation to the intersection between two functions

   x = 1.984/ln(x)

   
   
   starting from x_0 = 2.400 at k = 0 by forming a convergent Fixed Point Iteration, other than Newton's method. Demonstrate that the fixed point convergence criterion is satisfied near x_0. Tabulate

   k	0	1	2	3
   x_k

   for k = 0 to 3 iterations.

   ---

   Answer: If g(x)=1.984/ln(x), then g'(x) = -1.984/(x*ln^2(x)) and |g'(2.4)|=|-1.079|>1 indicating a marginal fixed point iteration divergence, which does not satisfy the problem statement. So, we take a manipulated inverse

   gi(x)=exp(1.984/x),   gi'(x) = -1.984*exp(1.984/x)/x^2   and   |gi'(2.4)|=|-0.7873|<1,

   
   
   satisfying the convergence criterion at x=2.4, with fixed point iterations x_{k+1}=gi_k, given in the table:

   k	0	1	2	3
   x_k	2.4	2.286	2.382	2.300

   Note that the original form converges as a fixed point iteration in spite of the marginal divergence warning and there are other forms that will work too.

4. Using the method of Golden Section Search, find the Maximum and its Interval of Uncertainty for

   g(x) = x*(2.718-e^(0.1234*x))

   
   
   on [3.6,6.0] at k = 1, say. Use the rounded version of the GSS constant (c =(4R) 0.3820). Summarize your results with a table of

   k	kfe	a_k	b_k	d_k	x_k	u_k	gx_k	gu_k
   1
   ....								......

   for k = 1 to 3 iterations with k_{fe} the number of function evaluations. Circle and label your best approximation to the maximum, its location, and state the interval of uncertainty.

   ---

   Answer: The tabulated answer is

   k	kfe	a_k	b_k	d_k	x_k	u_k	gx_k	gu_k
   1	2	3.6	6.0	2.400	4.517	5.083	4.390	4.298
   2	3	3.600	5.083	1.483	4.167	4.517	4.357	4.390
   3	4	4.167	5.083	0.916	4.517	4.733	4.390	4.377

   Hence, the Max[g(x)]=(4R)4.390, ArgMax[g(x)]=(4R)4.517 is the location, and IntervalOfUncertainy=(4R)[4.167,5.083].

In the following older problems prior to Fall 1999, use "CHOPPING EXAM PRECISION": The answers are calculated for chopping to 4 significant (4C) digits since the problems are from a time when chopping was used.

Note: Maple comments are not part of these sample exam problems, but were added afterward in the editing stage to aid in analyzing the problems.

1. Using the method of BISECTION starting with A(1)=1. and B(1)=2., find the root of F(X)=EXP(X)-3.5/X Record 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. Compare your answer to that using fsolve of Maple.

2. Find the root of F(X)=EXP(X)-3.5/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. Compare your answer to the Bisection and Maple answers from the first question.

3. Find the root of F(X)=EXP(X)-3.5/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. Compare your answer to the Bisection, Secant and Maple answers from the first and second questions.

4. 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. (Corrected (2/14/99) Final Ans.=(4ch) 1.998 using g(x)=exp(1/x)). Compare your answer to that using fsolve of Maple. Use the plot function of Maple to plot the problem function G(X).

5. Find the minimum of G(x)=EXP(x)+7.8/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. Use Maple's plot to plot the function G(X) and compare your problem answer to that using minimize of Maple.
   {CAUTION!: The Maple minimize function has a problem with function arguments having reciprocal terms like "1/x". See the example help page for minimize Minimizng a Function with Maple minimize (MapleVrel5 WorkSheet) or (MapleVrel3 HTML).

6. 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.: (0.8720,0.5609) for (X,G) or (U,G).) {WARNING!: You can Ignore the Maple part of this exercise since "maximize" obviously does not work for this simple trigonometric function. "Maximize" seems to work primarily for polynomial functions over algebraic fields. However, it may work if you approximate "x*cos(x)" by the first few terms of its Taylor series. Use Maple's plot to plot the function G(X) and compare your problem answer to that using maximize of Maple. You can also look for the critical point of the derivative "cos(X)-X*sin(x)" by using fsolve.}

7. Using Maple, get all the roots, including double and triple roots, of the polynomial
   
   x^5-11x^4+46x^3-90x^2+81x-27.

   Also plot the polynomial using the plot function of Maple, on [0,4].