Fixed Point Theorem and Sequences in One or Two Dimensions
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1 Fied Point Theorem and Sequences in One or Two Dimensions Dr. Wei-Chi Yang Let us consider a recursive sequence of n+ = n + sin n and the initial value can be an real number. Then we would like to ask if the sequence f n g converges or not. We eplore the following possibilities b starting with various initial values of : The rst number of each column below is the initial value for n+ = n + sin n : Eploration We observe the following iterations b using di erent initial values: For eample, if we start with = ; the table suggests the sequence converges to : If = ; we note that the sequence seems to converge to 9:8, which is close to : : 8 : 8 : : : 8 : : 8 : 9 : : 8 : 9 8 : 8 : 9 : : 9 : 9 : : : 9 8 : : 9 9: 9: 8 9: 8 9: 8 : : 8 : 9 8 : 8 : 9 : 8 : 8 : : : : 8 : 9 : 8 : 8 : : : 8
2 It is interesting to see if we start with = :8; the sequence seems to converge to :Net, did ou notice that if an iteration suggests a convergence, the limit would have to lie at the intersection of = and = + sin, which we sketch both graphs below: 8 8 Figure. Graph of = and = + sin :. Observations The sequence f n g will converge to either ; (k + ); or on the initial values for the sequence. Remarks: (k + ); where k = ; ; ::; depending. Wh does the sequence f n g not converge to k (where k is a positive or negative integer) even though the intersections of = and = f() = + sin show the intersections at ever k:. How do we formulate a proper theorem relating a sequence and a function?. Can ou recall how we show theoreticall a sequence f n g converges?. De nition of a Fied Point If f is a function from a set A into a set B then a point a in the set A is said to be a ed point of the function f if f (a) = a: Thus, if A is a set of real numbers then a number a is a ed point of f when the graph = f () meets the line = at = a:. Fied Point Theorem Suppose that f : [a; b]! [a; b] is a continuous and satisfing jf ()j < for each [a; b]: Then the function f can have at most one ed point in [a; b]: Eample We consider the function f de ned b f () = + cos
3 for ever [; ] : Observe that f : [; ]! [; ] : Now since jf ()j = sin for each we know that f is a contraction on the interval [; ] and therefore f must have a unique ed point in this interval.. Newton s Method Let f be a di erentiable function on (a; b):choose a point near a root r of f: De ne If lim n! n eists, then lim n! n = r: Eample Eercises for Newton s method: n+ = n f( n ) f ( n ) Problem How do ou derive the Newton s algorithm? Problem Describe how ou can solve a ed point problem b using the Newton s Method. Problem Describe how ou can turn a Newton s Method into a Fied Point problem.. Theorems. If n+ = f( n ) converges, where f : [a; b]! [a; b] is a continuous function, then the limit will be one of the solutions of f() = :. Let f : [a; b]! [a; b] be continuous, then the function f must have at least one ed point of [a; b]:. Let f : [a; b]! [a; b] be continuous and <. If jf ()j : Then the function f can have at most one ed point in [a; b]:. If n+ = f( n ) converges, and f satis es conditions above, then lim n! n is the solution of f() = : Remark: The theorems above can be generalized to higher dimensions. We will omit the statements here but we refer readers to the section of two dimensional eamples below.
4 . More one dimensional eamples Eercise. Discuss the convergence or divergence of the sequence n+ = f( n ); where f() = u( ); and u satis es one of the following three cases. Case. If u < : Case. If u = Case. If u > : Eercise. Discuss the convergence or divergence of the sequence n+ = f( n ); where f() = c ; and c satis es one of the following three cases. Case. If c > : Case. If c = Case. If c < : Eercise. Consider the sequence de ned b the function f() = arctan : Two dimensional eamples. Let f(; ) = ( + cos ; + cos( + )): De ne n+ = + cos n n ; n+ = + cos( n + n ); ( ; ) = (; ): Then prove that f(; ) = (; ) has a unique solution. Find lim n! ( n ; n ):.. A Contraction Function of Two Variables f (; ) = + cos ; + cos ( + ) for and : We write the coordinates of f (; ) as f (; ) and f (; ) for each point (; ) in the square [; ] [; ] : We observe that = sin sin sin ( + ) : sin ( Now point at sin sin sin ( + ) sin ( + ) and click on Evaluate. The norm is shown as the larger of two identical epressions and is therefore equal to either one of them. To obtain an upper bound for the norm of the matri as the point (; ) varies through the square [; ] [; ] we need to maimize the function g de ned b
5 v sin + sin s + sin ( + ) + g (; ) = ub t@ sin + sin C + sin + sin ( + ) + 8 sin sin ( + ) A: Point at this equation and click on De ne and New De nition and then click on Plot D and Rectangular. Open the plot properties dialog bo and set the and intervals to be [; ] : Then drag the epression into the sketch in order to add the plot z = to what we alread have. After ou have rotated the plot suitabl it will appear as follows, and we see from this gure that the maimum value of the function g is less than : We can therefore deduce from Subsection?? that the function f is a contraction. To show that the ed point of this function is approimatel (:89; :8), point at the sstem of equations + cos = + cos ( + ) = and click on Solve and Numeric. Finall, we shall approimate this ed point b iterating the function f starting at an arbitrar point in the square [; ] [; ] : Unfortunatel, Scienti c Notebook does not provide automatic iteration of a function of two variables but we can still iterate the function b
6 hand. We see that f (; ) = (: ; : ) f (: ; : ) = (: 9; : ) f (: 9; : ) = (: ; : 988) f (: ; : 988) = (: 9; : 9) f (: 9; : 9) = (: ; : 89) f (: 9; : 9) = (: ; : 89) f (: ; : 89) = (: ; : 8) f (: ; : 8) = (: 9; : 89) f (: 9; : 89) = (: ; : 88) f (: ; : 88) = (: 8; : 8) f (: 8; : 8) = (: ; : 89) f (: ; : 89) = (: 8; : 8) f (: 8; : 8) = (: 8; : 8) f (: 8; : 8) = (: 9; : 8) f (: 9; : 8) = (: 8; : 88) f (: 8; : 88) = (: 9; : 8) f (: 9; : 8) = (: 88; : 89) f (: 88; : 89) = (: 89; : 8) f (: 89; : 8) = (: 88; : 89) f (: 88; : 89) = (: 89; : 8) f (: 89; : 8) = (: 88; : 898) and this arra of iterates gives us about the same approimation to the ed point.
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