FUNCTIONAL EQUATIONS - University of California, Irvine

1 FUNCTIONAL EQUATIONSZHIQIN LU1. What is a FUNCTIONAL equationAn equation contains an unknown function is called a FUNCTIONAL following EQUATIONS can be regarded as FUNCTIONAL equationsf(x)= f( x),odd functionf(x)=f( x),even functionf(x+a)=f(x),periodic function, ifa,0 Example +1=an+an 1defines a FUNCTIONAL equation with the domain of which being nonnegative can also represent the sequence isf(n+1)=f(n)+f(n 1).Example (Radioactive decay) Letf(x) represent a measurement of the numberof a specific type of radioactive nuclei in a sample of material at a given assume that initially, there is 1 gram of the sample, that is,f(0)=1. By thephysical law, we havef(x)f(y)=f(x+y).Can we determine which function this is?

2 2. Substitution methodExample ,1. Solve the equationa f(x)+f(1x)=ax,where the domain offis the set of all non-zero real : November 7, notes for theMath Circle, Irvine . Partially supported by the NSF grant LU, DEPARTMENT OF MATHEMATICSS olution:Replacingxbyx 1, we geta f(1x)+f(x)= therefore have(a2 1)f(x)=a2x ax,and hencef(x)=a2x axa2 1. Exercise the FUNCTIONAL equation (a2,b2)a f(x 1)+b f(1 x)= EQUATIONS3 Exercise a functionf:R\{0} Rsuch that(1+f(x 1))(f(x) (f(x)) 1)=(x a)(1 ax)x,wherea (0,1).3. RecurrenceRelationsExample (Fibonacci EQUATIONS ) Letf(n+2)=f(n+1)+f(n)withf(0)=0,f(1)=1. Find a general formula for the :We consider the solution of the formf(n)= nfor some real number . Then we have n+2= n+1+ nfrom which we conclude that 2= +1.

3 Therefore 1=1+ 52, 2=1 LU, DEPARTMENT OF MATHEMATICSA general solution of the sequence can be written asf(n)=c1 1+ 52 n+c2 1 52 n,wherec1,c2are coefficients determined by the initial values. By the initial condi-tions, we havec1+c2=0c1 1+ 52 +c2 1 52 =1 Thus we havec1=1 5,c2= 1 (n)=1 5 1+ 52 n 1 52 n .It is interesting to see that the above expression provide all positive integers for anyn. Exercise the sequence defined byan=3an 1 2an 2forn 2 with the initial conditiona0=0,a1= EQUATIONS54. TheCauchy sMethodExample thatfis a continuous function onR. Assume that for anyx,y Rf(x+y)=f(x)+f(y).Find the functionf(x).Solution:First, we havef(0)=0. Letc=f(1).Using the math induction, we havef(x1+ +xn)=f(x1)+ +f(xn).

4 Letx1= =xn=x. Then we getf(nx)=n f(x)for any positive integern. Letx=1/mwheremis a nonzero integer. Then we havef(nm)=n f(1m).On the other hand,m f(1m)=f(1)= we havef(nm)=n f(1m)= conclusion here is that for anyrationalnumber , we havef( )=c .Iffis continuous, then we conclude that for any real numberx,f(x)=cx=x f(1).For those of you who are not familiar with the concept of continuity, the as-sumption can be weakened to the boundedness of the function. Assume thatfisbounded. Letxbe any real number. For any >0, we choose a rational number such that|x |< . LetNbe the integer part of the 1/ . Then|f(N(x ))| Cbecause the function is bounded. Thus we have|f(x) f(1) |=|f(x) f( )| we choose to be so small, then we must havef(x)=f(1)x.

5 6 ZHIQIN LU, DEPARTMENT OF MATHEMATICSE xercise (Radioactive Decay) Solvef(x+y)=f(x)f(y),wherefis Using FUNCTIONAL equation to define elementary functionsOne of the applications of FUNCTIONAL EQUATIONS is that they can be used to char-acterizing the elementary functions. In the following, you are provided exercisesfor the FUNCTIONAL EQUATIONS for the functionsax,logax, tanx, sinx, and cosx. Canyou setup the FUNCTIONAL EQUATIONS for cotx, secx,cscx, and hyper-trigonometricfunctions?Exercise logarithmic function satisfies the propertylog(xy)=logx+logyfor any positive real numbers. Solve the equationf(xy)=f(x)+f(y),where bothx,yare positive real numbers, wherefis EQUATIONS7 Exercise the equationf(x+y)=f(x)f(y)wherex,yare any real numbers, wherefis the equationf(x+y)=f(x)+f(y)1 f(x)f(y)for any real numbersx,y, wherefis LU, DEPARTMENT OF MATHEMATICSE xercise the EQUATIONS for continuous/bounded functionsf(x),g(x)f(x+y)=f(x)g(y)+f(y)g( x)g(x+y)=g(x)g(y) f(x)f(y)for any real numbersx, EQUATIONS96.

6 AdditionalProblemsExercise thata0=1,a1=2, andan=4an 1 an 2forn 2. Find a prime factor problem is, obviously, obsolete. By th year 2016 is also special. It is thedimensional of the space of all 26 26skew-symmetric matrices. Just (x)+f(x 1x)=1+x,x,0,1 Exercise revisited: is the solution unique?Exercise we drop the assumption thatfis continuous or boundedness inExample , is the functionfstill linear?10 ZHIQIN LU, DEPARTMENT OF MATHEMATICS7. Solutions of theExercisesSolution of Exercise :Replacingx 1 byx, we obtaina f(x)+b f( x)=c(x+1).In the above equation, if we replacexby x, we geta f( x)+b f(x)=c( x+1).Thus we solvef(x)=ac(x+1) bc( x+1)a2 verification, the above is indeed the solution. Solution of Exercise :By replacingxwithx 1, we obtain(1+f(x))(f(x 1) (f(x 1)) 1)=(x a)(1 ax)x=(1+f(x 1))(f(x) (f(x)) 1).

7 Thus we havef(x)=f(x 1),and hencefis determined by the equation(1+f(x))(f(x) (f(x)) 1)=(x a)(1 ax)x. Solution of Exercise :an=2n 1. Solution of Exercise :We first conclude thatf(x1+ +xn)=f(x1) f(xn).By taking allxito be the same, we obtainf(nx)=(f(x)) particular,f(1m)=(f(1)) therefore we havef(nm)=(f(1)) continuity/boundedness ensures that the above equation is also valid for irra-tional numbers. FUNCTIONAL EQUATIONS11 Solution of Exercise :Letg(x)=f(ex). Theng(x+y)=f(ex ey)=f(ex)+f(ey)=g(x)+g(y).The problem is reduced to Example Solution of Exercise :Eitherf(x) 0 orf(x) is positive. So we can takeg(x)=logf(x) in casef(x) is positive. Then we reduce to Example Solution of Exercise :Takeg(x)=arctanf(x).

8 Solution of Exercise :Takeh(x)=g(x)+ 1f(x). Then we obtainh(x+y)=h(x)h(y).Then using Exercise 8. Solutions of the additional problemsSolution of Exercise :an=12((2+ 3)n+(2 3)n).Since(2+ 3)5n+(2 3)5n=((2+ 3)5+(2 3)5) integer,and since (2+ 3)5+(2 3)5=724=181,a2015is divisible by the primenumber 181. Solution of Exercise :Replacingxbyx 1x, we obtainf(x 1x)+f( 1x 1)=1+x 1x 1, we obtainf( 1x 1)+f(x)=1 1x we havef(x)=12(1+x+1 1x 1 1 x 1x)=12(x+1x 1x 1). Solution of Exercise :Yes, the solution is unique, even in general a cubic equa-tion has three real is because ifxis small, the cubic equation only has one real LU, DEPARTMENT OF MATHEMATICS Solution of Exercise :The set of real numbersRcan be considered as a vectorspace over rational numbersQ.

9 A basis for such a vector space is called aHamelbasis. That is, there is a set of real numbersu such that for any real numbers ,there is a unique representation of the form =N i=1riu existence of such a basis is equivalent to theAxiom of Choice. Without assum-ing the continuity or boundedness, we can still prove thatfis alinear transforma-tionwhen regardingRas a vector space overQ. However, one can prescribe anyvalue on the Hamel basis to construct a (unbounded) solution. So in conclusion,the solution is not unique without the additional assumption. Professor, Department ofMathematics, University ofCaliforniaIrvine,E-mail