Processing math: 100%

Spring 2016, problem 11

Consider a function f:QQ which satisfies 3f(w+x+y+z)+f(w+x)+f(x+y)+f(y+z)+f(z+w)+f(w+y)+f(x+z) =2f(w+x+y)+2f(x+y+z)+2f(y+z+w)+2f(z+w+x). Find all possible such functions f(x).

Comments

Nelix
5 years ago

The possible functions are exactly those of the form f(x)=αx2+βx with α,βQ.

Proof:

Call the equation from the problem statement "the PE". The functions g(x)=x and h(x)=x2 are easily checked to solve the PE. Since the PE is linear, every linear combination f(x):=αh(x)+βg(x)=αx2+βx will solve the PE. It remains to be proved, that all solutions are of this form.

So let f be any solution of the PE. Letting w=x=y=z=0 in the PE gives us f(0)=0. Letting (w,x,y,z)=(a,a,a,(z3)a) for some zZ and aQ gives:

f(za)=13(2f(3a)3f(2a))+2f((z1)a)f((z2)a. [Eq.1]

Now we keep a0 fixed. We define dz:=f(za)f((z1)a) and C:=23f(3a)f(2a) and rewrite [Eq.1] as:

dz=dz1+C. The general solution to this recursion is dz=Cz+q with qQ. But, since f(0)=0, we have:

f(nx)=ni=1di=C2n(n+1)+qn=C2n2+(C2+q)n for every nN [Eq.2].

The solution [Eq.2] of the recursion [Eq. 1] can be extended routinely back to negative n. Define α:=C2 and β:=C2+q. We rewrite:

f(za)=αz2+βz for all zZ.

Since a was arbitrary, for a given rN there exist α1 and β1, such that:

f(zar)=α1z2+β1z.

Now pick an sZ and let z:=sr. Then we have: f(sa)=α1r2s2+β1rs=αs2+βs.

Comparing coefficients: α1=αr2 and β1=βr. So:

f(zra)=αa2[azr]2+βaazr.

But as we go through all zZ and rN the number azr goes through all rationals. Therefore:

f(q)=αa2q2+βaq for all qQ, as desired.

Call the equation from the problem statement "the PE". The functions g(x)=x and h(x)=x2 are easily checked to solve the PE. |

arlensurrealistic 1 month ago
pkornya
5 years ago

WecalculatesuccessivetermsofNewtonsforwarddifferencingformulaf(x)=f(0)+n=1(xn)Δnf(0)Supposethatf:QQisafunctionsatisfyingtheconstraintintheproblem.Letw=x=y=z=0.Then9f(0)=8f(0).Thereforef(0)=0.LetxQbearbitraryandletw=0,y=z=1.Substitutingintotheconstraint3f(x+2)+f(x)+f(x+1)+f(2)+f(1)+f(1)+f(x+1)=2f(x+1)+2f(x+2)+2f(2)+2f(x+1)andsolvingforf(x+2)f(x+2)=2f(x+1)f(x)+f(2)2f(1)ThereforeΔ2f(x)=f(x+2)2f(x+1)+f(x)=f(2)2f(1)SincexQisarbitraryΔ2f(x+1)=Δ2f(x),ThenΔ3f(x)=Δ2f(x+1)Δ2f(x)=0Byinductiononn3Δn+1f(x)=Δnf(x+1)Δnf(x)=0.InparticularΔnf(0)=0foralln.SubstitutingintoNewtonsforwarddifferenceformulaf(x)=f(1)(x1)+[f(2)2f(1)](x2)=αx+βx2whereα,βQ.Converselyiff(x)=αx+βx2withα,βQthenitisroutinelyverifiedthatfsatisfiestheconstraint.

IkramaKhan
1 month ago

f(x)=−x 2 +6x−8 We can see function is a polynomial and the domain of polynomial function is real number. ∴ x∈R f(x)=−x 2 +6x−8 =−(x 2 −6x+8) =−(x 2 −6x+9−1) =−(x−3) 2 +1 Maximum value of −(x−3) 2 would be 0 ∴ Maximum value of −(x−3) 2 +1 would be 1. ∴ f(x)∈(−∞,1] We can see from the given graph that function is symmetrical about x=3 and the given function is bijective. So, x would be either (−∞,3] or [3,∞) The correct option which satisfy A and B both is: A=(−∞,3] and B=(−∞,1]

IkramaKhan
1 month ago

In this question the origin is taken to be at a harbour and the unit vectors i and j to have lengths of 1 km in the directions E and N. A cargo vessel leaves the harbour and its position vector t hours later is given by r1 = 12ti + 16tj. A fishing boat is trawling nearby and its position at time t is given by r2 = (10 - 3t)i + (8 + 4t)j. Here is the

cartwrighthlmv67
1 month ago

The general solution to this recursion is dz=Cz+q with q∈Q. But, since f(0)=0, we have:

f(nx)=∑ni=1di=C2n(n+1)+qn=C2n2+(C2+q)n for every n∈N [Eq.2].