5.5 Newton s Approximation Method

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1 498CHAPTER 5. USING DERIVATIVES TO ANALYZE FUNCTIONS; FURTHER APPLICATIONS 4 3 y = x 4 3 f(x) = x cosx y = cosx 3 3 x = cosx x cosx = 0 Figure 5.: Figure showing the existence of a solution of x = cos x which occurs where the graphs of y = x and y = cos x intersect, or equivalently a solution of f(x) = 0 where f(x) = x cos x. 5.5 Newton s Approximation Method While there are many types of equations we can solve algebraically and students typically spend much time in their young years learning to solve them there are numerous equations we cannot solve algebraically. We can often demonstrate that there exist solutions, and perhaps even discover the number of solutions, without actually knowing the exact solutions. Within industry approximate solutions often suffice, though the level of precision needed varies from problem to problem. When the problem is very complex, or an approximate solution is needed quickly, efficiency becomes an important consideration. Efficiency, accuracy, and prospects for success are all factors when choosing a mehtod, and what works well for one problem might not work well or even at all for another. In this section we will look at an approximating scheme attributed to Newton, though the modern interpretation has been ascribed to Thomas Simpson (of Simpson s Rule, Section 8.3). We will also examine other methods for comparison Approximating Methods Applied to an Example Consider the equation x = cosx. (5.0) We can tell from the first figure of Figure 5. that there is a solution of this equation somewhere in x [0, ]. In fact, by the Intermediate Value Theorem (page 96) we can look at f(x) = x cosx (5.) and note that f(x) is continuous everywhere, f(0) = 0 = < 0, while f(π/) = π/ 0 = π/ > 0, so there is a solution of f(x) = 0 in x [0, π/], meaning there is a solution of

2 5.5. NEWTON S APPROXIMATION METHOD 499 x = cos x there as well. Rather than solving the original equation (5.0), we will instead solve the equivalent equation (5.). To do so, we will consider three methods in turn. Solve by graphing: If we have access to a device such as a graphing calculator, we can make successive magnifications of the window near the apparent solution, and thereby zoom in to the solution repeatedly, in the meantime reading off better and better approximations from the provided axes or curve tracing features of the device. While visually appealing, it is not an easy process to train a primitive device to implement. Method of Bisection: This method is based on the Intermediate Value Theorem, where we check the sign of f(x) at various points, and once we detect a sign change on an interval, we check the sign of f(x) at the midpoint of the interval to see which subinterval (left or right) contained the sign change, and then we continue, each time halving the length of the interval with the sign change. In this way we can ultimately get as close to the actual value as we wish. For instance, if we wish to attempt this method with our current problem, we would compute } f(0) = < 0 sign change in [0, ]. f() = cos > 0 Next we consider the value of f(x) at the midpoint of [0, ], and include the previous values for clarity: f(0) = < 0 f() = cos > 0 sign change in [0, ]. f() = cos > 0 Next we compute f(x) at the midpoint of [0, ], again including the two endpoints for clarity: f(0) = < 0 f(0.5) = 0.5 cos < 0 sign change in [0.5, ]. f() = cos > 0 Next we compute f(x) and the midpoint of [0.5, ] and compare it to the endpoints: f(0.5) = 0.5 cos < 0 f(0.75) = 0.75 cos > 0, sign change in [0.5, 0.75]. f() = cos > 0 Next we compute f(x) and the midpoint of [0.5, 0.75], namely ( ) = 0.65: f(0.5) = 0.5 cos < 0 f(0.65) = 0.65 cos < 0 sign change in [0.65, 0.75]. f(0.75) = 0.75 cos > 0 This tells us the solution is somewhere in [0.65, 0.75], whose midpoint is ( ) = , and so on. If we take x 0 = 0, x =, and then consider the midpoints of the relevant intervals generated by the sign changes to be x =, x 3 = 0.5, x 4 = 0.75, x 5 = 0.65, and so on, we see a sequence of x-values which get progressively closer to the actual solution. A sign chart for the f(x n ) values helps to illustrate this phenomenon:

3 500CHAPTER 5. USING DERIVATIVES TO ANALYZE FUNCTIONS; FURTHER APPLICATIONS sign f(x n ): x = x = x4 = 0.75 x5 = 0.65 x3 = 0.5 x0 = 0 While this method can be easily programmed into a computer or similar device, it is not usually the most efficient computationally, meaning it takes more steps to get the same accuracy as the next method, which takes some advantage of the actual slope of the function, i.e., the function s derivative. Newton s Method: With Newton s method, we begin with one guess, which we will call x. We then look at the tangent line to f(x) at (x, f(x )), the slope of this line being f (x ). We then take x to be the x-intercept of this tangent line (which would be a solution if the curve s slope were constant!), in essence following the tangent line to the x-axis to find x. We then take the tangent line at (x, f(x )) and follow it to the x-axis to find x 3, and follow the tangent line at (x 3, f(x 3 )) to the x-axis to find x 4, and so on, to produce a sequence x, x, x 3,, which will often converge very quickly towards a point x which solves f(x) = 0. This is illustrated for our example above, namely solving x = cosx, i.,e., f(x) = x cosx = 0 using x = 0, and the x n following for n =, 3, 4,. This is illustrated in Figure 5.. Algebraically, for Newton s Method we let x n be our nth guess, and then its tangent line to y = f(x) at x = x n is given by y = f(x n ) + f (x n )(x x n ), which when we set equal to zero to find the x-intercept giving us f(x n ) + f (x n )(x x n ) = 0 f (x n )(x x n ) = f(x n ) x x n = f(x n) f (x n ) x = x n f(x n) f (x n ). Taking x n+ to be this x-value at which the tangent line through (x n, f(x n )) intersects the x-axis, we get the following, recursive formula: x n+ = x n f(x n) f (x n ). (5.) For our present example in which f(x) = x cosx and f (x) = +sinx, our recursion relationship becomes x n+ = x n x cosx + sinx. So far we have considered x = 0 to be our first guess. Below we list values of x n where we begin first with x = 0, and then also consider x =, x = and x = 0. Note that approximations are given to ten significant digits.

4 5.5. NEWTON S APPROXIMATION METHOD 50 (x, f(x )) y = f(x) x x 3 x (x, f(x )) Figure 5.: Figure showing two iterations of Newton s Method for finding approximate solutions of f(x) = x cos x = 0, i.e., solutions of x = cos x, with an initial guess of x = 0. n x n n x n n x n For this particular example, it is clear that a good first guess x can cause the algorithm to converge quickly to a stable value. However, using x = 4 here leads to unstable oscillation in the sequence {x n }, and no apparent convergence. The behavior is not always predictable at first glance (the reader is invited to note that x = actually converges faster to the stable value of x n than does letting x = 0, though the solution is closer to 0 than ). To be sure we have an approximate solution, one can check cos , as desired. Note that when the method does converge, it often does so quite quickly: we managed to get answers accurate to 0 significant digits after 5 6 iterations. To use our method of bisection and have accuracy of ±0 0 we would need to bisect the interval [0, ] some n times, giving a subinterval of length / n where / n 0 0, i.e., n 0 0, which means ( n)log 0, or (n )log 0, or n + 0/ log 34., so we would need 35 iterations of the Method of Bisection to achieve the same accuracy for this problem. That is certainly not a problem for a computer program to accomplish quickly, but the 5 6 steps required

5 50CHAPTER 5. USING DERIVATIVES TO ANALYZE FUNCTIONS; FURTHER APPLICATIONS here show Newton s Method to be more efficient for this particular problem. 5 In the typical application of Newton s Method, one attempts to solve an equation that has been put into the form f(x) = 0, makes a guess for x, runs the algorithm 6 and inspects for convergence. If there will be convergence, it is usually readily apparent. If not, that is usually clear as well and another attempt to choose an x that will cause convergence. There are some functions which will not allow for convergence, and a method such as the Method of Bisection can be attempted. (We will consider such situations later.) The electronic ability to graph the function in question can be very helpful as well, even if only to determine where to begin whichever method is chosen. It is also the case that there can be several solutions of f(x) = 0, and perhaps some graphical analysis needs to be done at first so we know how many solutions exist, and approximately where they are located. Example 5.5. Find all real solutions of x 3 = 3x. Solution: We begin as before by making this a question about zeros of a function f(x) = x 3 +3x. Assuming we had no graph to work with, we could next seek out intervals on which the function changes signs: x f(x) We see sign changes in [ 3, ], [, 0], and [0, ]. We know from Algebra that there can be at most three solutions of f(x) = 0 for the case that f(x) is a third-degree polynomial, so we need only look for approximations of the solutions in these three intervals. We will choose x to be 5 Indeed such problems were and still are typically solved by students using pencil-and-paper and very simple scientific calculators in homework and exam settings for decades. 6 Three common ways to run the algorithm are through the use of a calculator, a computer programming language, or a spreadsheet. With most graphing calculators today, we can use the ANS and ENTER keys to run the recursive steps easily. For instance, if we type our x value 0 ENTER, the display shows 0 as our answer. We then can type ANS-(ANS-(COS(ANS))/(+SIN(ANS)) ENTER and the calculator will return the output from entering our first answer (x = 0) after it is run through the right-hand side of (5.), namely x. At that point, x is our new answer, so if we type ENTER again it will repeat the line above with ANS= x, giving our new answer, namely x 3 and so on. Repeatedly pressing ENTER then outputs x 4, x 5 and so on. On a spreadsheet, one can enter 0 into cell A, and then in A enter = A (A cos(a))/( + sin(a)), causing the contents of A to be as above. One can then copy cell A and paste its contents simultaneously into A3, A4 and so on, and the spreadsheet will likely understand this action to mean each cell should reference the one above using a similar formula, i.e., so that A = A (A cos(a))/( + sin(a)), A3 = A (A cos(a))/( + sin(a)), A4 = A3 (A3 cos(a3))/( + sin(a3)), and so on. This will in fact make it easier to try new first guesses for A (think x ), since editing A to be a different number will automatically update A, A3 and so on. Using either the programmable features of a graphing calculator, or an actual computer programming language are alternative strategies, as is simply using calculator memory (or re-typing) and processing the numbers by brute force. Of course, regardless of the chosen method, we have to be sure that the calculating device is reading angles in radians.

6 5.5. NEWTON S APPROXIMATION METHOD Figure 5.3: Graph of f(x) = x 3 + 3x for Example Note the intervals on which there is a sign change, and the approximate placements of the solutions of f(x) = 0, i.e., where x 3 = 3x. the midpoint in each case. We will use these in our recursive formula, which for this problem becomes: x n+ = x n f(x n) f (x n ) = x n+ = x n x3 + 3x 3x + 6x. Using x =.5, 0.5, 0.5 respectively we generate the following tables: x n n x n n x n n We conclude that approximate solutions of x 3 = 3x are x , and The graph of f(x) = x 3 + 3x is given in Figure 5.3. From both the geometric interpretation (of following the tangent lines back to the x-axis) and the recursion formula that a flatter curve, i.e., smaller f (x n ) will cause x n+ to be more distant from x n. In the above example, at x = 0 we would have f (x) = 0 and therefore the tangent line would never intercept the x-axis. We would also find ourselves with a zero denominator in the recursion formula. A common example showing graphically how the x n can diverge is given next. Example 5.5. Suppose we wish to use Newton s Method to solve 3 x = 0. While we can see the solution is x = 0, instead we will suppose we do not know this, and instead make our first

7 504CHAPTER 5. USING DERIVATIVES TO ANALYZE FUNCTIONS; FURTHER APPLICATIONS x 4 x x x Figure 5.4: Figure showing the tangent lines producing the succession of the x, x, x 3, generated by the Newton s Method recursion formula for f(x) = 3 x = 0 if x =, as in Example As the tangent lines become more horizontal, the x n values diverge from each other. guess of x =. Our recursion relation becomes Running our algorithm, we get x n+ = x n f(x n) f (x n ) = x n x/3 3 x/3 = x n 3x n. n x n and so on. Indeed we can see a simple pattern where x n+ = x n, and the x n diverge. This coincides with the slopes of the tangent lines shrinking in size, the effect of which is shown graphically in Figure 5.4. As has been pointed out, when the slope of the curve decreases in size, the method is likely to produce points x n which are more spread apart, and therefore less likely convergent. However, the method can be quite useful when algebraic methods are not up to the task. It is helpful to have a general idea of the behavior of the function s graph, such as the number of solutions.

8 5.5. NEWTON S APPROXIMATION METHOD 505 Exercises. Newton s Method produces a method to compute square roots which was used for many years by students who had no access to calculators. For instance, to compute 3, one would look at the function f(x) = x 3 and use the method to find where f(x) = 0. In this way, use Newton s Method to compute 3, 4 and 36 accurate to ±0 9 or better.. As above, use Newton s Method to compute 3 9 accurate to ±0 9 or better. 3. Consider the equation tan x = x. (a) Sketch rough graphs of y = tanx and y = x. Note that x = 0 is clearly one solution. (b) Use Newton s Method to find two other positive solutions. In doing so, note what happens if we attempt the method with various initial values x = 4., 4., 4.3, 4.4, 4.6, 4.7, 4.8 and 4.9. Do so again with x ranging from 7.3 to Using only a scientific (non-graphing) calculator or similar device, graph f(x) = x 3 + x 5x. To do so: (a) Find the critical points (using the Quadratic Formula) and produce a sign chart for f (x). Use this to identify local extrema. (b) Find any inflection points and produce a sign chart for f (x). (c) Use Newton s Method to find all (three) x-intercepts. (d) Sketch the graph, reflecting all of this information, along with the y- intercept.