Lecture 07. (with solutions) Summer 2001 Victor Adamchik. 2 as an infinite product of nested square roots. M #'((((((((((((((((((((((((((((( 2 ] # N

Transcription

1 Lecture 07 (with solutions) Summer 00 Victor Adamchik Off#General::spell'; Off#General::spell'; $Line 0; Exercise Vieta's formula, developed in 593, expresses cccc as an infinite product of nested square roots. S ccc S m, ccc #r$$$ 0#- ccc #%&&&&&&&&&&&&&&&&. r$$$ L # ccc M #'(((((((((((((((((((((((((((((. %&&&&&&&&&&&&&&&&. r$$$ \ ] # N ^ Implement this in Mathematica using Nest and Fold. Ã Solution VietaSequence#n_' : cccc #Rest$NestList$r$$$$$$$$$$$. # &,0,n(( Times VietaSequence#' ssn s Pi ss N Why is Vieta's formula correct? We know that sin+/ cos- ccccc sin- ccccc We iterate this formula in Sin: sin- ccccc cos- ccccc 4 sin- ccccc 4 sin+/ cos- ccccc sin- ccccc cos- ccccc cos- ccccc 4 sin- ccccc 4

2 lecture07post.nb After n steps, we have sin+/ n cos- ccccc cos- ccccc 4...#cos- ccccccc sin- ccccccc n n We divide both sides by, and find a limit when n ˆ: sin+/ cccccccc ccccccc cos- ccccc cos- ccccc 4...#cos- ccccccc sin+ ccccc / cccccccccccccccc n cccccc n ccccc n It follows then sin+/ cccccccc ccccccc ˆ Ç k cos- ccccccc k We substitute S cccc : ccccc S ˆ Ç k S cos- ccccccccccccc k. All that is left to do is to verify that the cosines are nested square roots. Mathematically this would be proved by induction, based on the formula: cos+/ '(((((((((((((((((((((((((((((((((( ccccc +cos+ /. / Exercise In 989 Borwein expressed S as the limit of a sequence defined by an iteration: lim n ˆ an ccccc S where a n +y n. / 4 a n n0/ y n +y n. y n. / a # r$$$ and

3 lecture07post.nb 3 y 0 %&&&&&&&&&&&&&&&&&& y n0 n cccccccccccccccccccccccccccccccc ccccc. %&&&&&&&&&&&&&&&&&& y n0 y r$$$ 0 0 Implement this in Mathematica using Nest and Fold. Ã Solution Clear#a, y, n'; y#n_' : Nest# #^4/^+ s 4// s #^4/^+ s 4// &, Sqrt#' 0, n' a#n_' : Fold# ++# +y##'. /^4/ 0 ^+3. #+# 0 // y##' - +y##'^. y##'. / &/, Sqrt#', Table#i, ;i,, n?'' The number of significant digits in iteration formula grows quadratically: $MaxExtraPrecision 00; TableForm#N#Table# s a#n' 0 S, ;n, 0, 3?', '' Fixed Point A point which does not change upon application of a map. Suppose F is a map, then x - such as x - F+x - / is a fixed point. Consider a continued fraction from the previous lecture. cccccccccccccccc ccccccccccc. ccccccccccccccccccccc ccccc. cccccccccccccccc cccccc. cccccccc cccccccc cccc... ccccc The fraction is defined by a recurrence relation The fixed point x +/ of this map occurs when Solving this quadratic equation, gives x n. cccccccc ccccc x n0 x +/. ccccccccccc x +/

4 4 lecture07post.nb Solve#x m. s x, x' <<x cccc r$$$, 0 5 0, <x cccc,. r$$$ 5 0 N#%' ;;x ?, ;x.6803?? In the previous lecture we found this point numerically NestList#. s # &,, 5' ss N ;.,.,.5,.66667,.6,.65,.6538,.6905,.6765,.688,.6798,.6806,.6803,.6804,.6803,.6803? As the alternative to Nest and NestList, Mathematica has FixedPoint function, which is very similar to Nest except that it continues to apply a given function until the result stop changes. Here we use FixedPoint to compute the above fraction. NOTE, we use a float. as a starting point FixedPoint#. s #&,.'.6803 otherwise, the iteration will never stop. FixedPoint uses SameQ (===) as the default test for convergence. However, no two of the rational terms are ever identical. Similar to NestList, FixedPointList returns a list of the intermediate results FixedPointList#. s #&,.' ;.,.,.5,.66667,.6,.65,.6538,.6905,.6765,.688,.6798,.6806,.6803,.6804,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803,.6803? Here is another example x n x %&&&&&&&&&&&&&&&&&&& n0 0 x n0 x 0 cccc FixedPointList###Sqrt# 0 #^' &, s ' < cccc r$$$ r$$$, 3 ccccccccc, 3 ccccccccc

5 lecture07post.nb 5 which shows how dangerous could be the last argument. In this example, the exact number cccc but a float won't work at all words just fine, FixedPointList###Sqrt# 0 #^' &, 0.5, 50' ;0.5, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ? FixedPoint accepts a third argument, which terminates the iteration after that particular number of steps. The another possibility would be to increase the precision of calculations: FixedPointList###Sqrt# 0 #^' &, N# s, 0'' ; , , ? Newton iteration We now use Newton's method to find the square root of the number. This involves solving f +x/ x 0 iteratively. The general Newton method is based on this iteration. For finding the square root, the iteration reduces to x i. x i 0 f +x i/ cccccccccccccccc f +x i / Here is the square root of S. x x i. i. cccccccc cccccccc cccc #x i FixedPoint#+#^. Pi/s+##/ &,.'.7745 Just 0 iterations give us 00 significant digits:

6 6 lecture07post.nb Sqrt#Pi' 0 FixedPointList#+#^. Pi/ s+##/ &, N#, 0'' ; g , g , g , g , g , g , g , , , ? Length#%' 0 Now we are going to apply the Newton iteration to the higher order polynomials. The question we are interested in is: given a starting value, to which root will the solution converge? First we define a function (according to the Newton method) Clear[iteration]; iteration[poly_, _] := Function[, Evaluate[Together[ - poly/d[poly, ]]]] where poly is any polynomial, and is a starting value (a complex number). We will pick from the unit square data Table#x. Iy,;x, 0.,., 0.07?, ;y, 0.,., 0.07?'; and find all fixed points (we restrict our search to 75 iterations) of the cubic: points#3' FixedPoint#iteration#^3 0, ', #, 75' & s To visualie the data in a form of a matrix we use ListDensityPlot: data;

7 lecture07post.nb 7 ListDensityPlot#Im#points#3'', Mesh False, ColorFunction Hue, MeshRange ;;0,?, ;0,??' h#densitygraphics#h Consider the quadratic:

8 8 lecture07post.nb points#4' FixedPoint#iteration#^4 0, ', #, 75' & s ListDensityPlot#Im#points#4'', Mesh False, ColorFunction Hue, MeshRange ;;0,?, ;0,??'; data; 0 - and the quintic:

9 lecture07post.nb 9 points#5' FixedPoint#iteration#^5., ', #, 75' & s ListDensityPlot#Im#points#5'', Mesh False, ColorFunction Hue, MeshRange ;;0,?, ;0,??'; data; These pictures show the convergence to the roots as a function of the complex start value. Mandelbrot Set The Mandelbrot set is the set of all complex numbers c for which sequence defined by the iteration x n. x n. c, x 0 c remains bounded. This means that there is a number B such that the absolute value of all iterates x n never gets larger than B. Let us experiment by setting c to different values: c ; NestList##^. c &, c, 4' ;,, 5, 6, 677? The sequence is growing and thus unbounded. We can say that c does not belong to the Mandelbrot set. c 0; NestList##^. c&, c, 4' ;0, 0, 0, 0, 0?

10 0 lecture07post.nb The sequence oscillates between - and 0, so it remains bounded (but it does not converge). You may ask, what's so special about this iteration. Much of the fascination of the Mandelbrot set is the fact that an extremely simple formula gives rise to an object of such great complexity. Let us implement the Mandelbrot set using functions FixedPoint and FixedPointList. c 0.; FixedPoint##. c &, c' 0.70 If c is the iteration does not converge. To stop the iteration, we specify the number of steps: c ; FixedPoint##. c &,c,0' g g Another tool to stop iteration is to use the option SameTest: c ; FixedPoint##. c &, c, SameTest +Abs## 0 #'! 0 &/' 6 The function Abs## 0 #'! 0 & inside of SameTest means that we stop iterations when the difference between two steps is greater than 0. Here is the Mathematica definition for the Mandelbrot function Mandelbrot#c_' : Length#FixedPointList##. c &,c,80, SameTest +Abs## 0 #'! 5. ««Abs## 0 #'? 0^06. &/'' Note, that we are interested in the number of iterations that it takes until a fixed point has been reached. That's why we wrapped Length around FixedPointList. We are using SameTest to terminate recursion if the distance between two iterations is bigger than 5. Also we set the limit of iterations to 80.

11 lecture07post.nb DensityPlot#0Mandelbrot#x. y I', ;x, 0.5, 0.5?, ;y, 0,?, Mesh False, Frame False, AspectRatio Automatic, PlotPoints 5'; Different shadows of gray correspondent to a different number of iterations. If you want to see a colorful picture, you add ColorFunction Hue to it:

12 lecture07post.nb DensityPlot#0Mandelbrot#x. y I', ;x, 0., 0.4?, ;y, 0.5, 0.65?, Mesh False, Frame False, AspectRatio Automatic, ColorFunction Hue, PlotPoints 00' h#densitygraphics#h Exercise 3 Find all fixed points of the second order of the Mandelbrot map. x n. x n. c Points which are invariant after two iterations. Ã Solution After two iterations: x n. x n.. c +x n. c/. c Solve# m +^. c/^. c, ' <; +0/ s3?, ; 0+0/ s3?, < cccc r$$$,0 0 Ç 7 0, < cccc,0. Ç r$$$ 7 0

13 lecture07post.nb 3 Solve# m ^. c, ' ;; +0/ s3?, ; 0+0/ s3?? c s ; cccc r$$$$$$$$$$$$$$$$$$, c0 r$$$ cccc,0 0 Ç 5 0 FixedPointList$#. c&, cccc. cccc Ç,5( < cccc. cccc Ç, cccc. cccc Ç FixedPoint$+#. c /. c&, cccc r$$$$,0 0 Ç 5 0, (ssn Ç r$$$$$$$$$$$$$$$$$$ cccc, c0ssN Ç