Spring 2016, problem 11
Comments
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. |
WecalculatesuccessivetermsofNewton′sforwarddifferencingformulaf(x)=f(0)+∞∑n=1(xn)Δnf(0)Supposethatf:Q→Qisafunctionsatisfyingtheconstraintintheproblem.Letw=x=y=z=0.Then9f(0)=8f(0).Thereforef(0)=0.Letx∈Qbearbitraryandletw=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)Sincex∈QisarbitraryΔ2f(x+1)=Δ2f(x),ThenΔ3f(x)=Δ2f(x+1)−Δ2f(x)=0Byinductiononn≥3Δn+1f(x)=Δnf(x+1)−Δnf(x)=0.InparticularΔnf(0)=0foralln.SubstitutingintoNewton′sforwarddifferenceformulaf(x)=f(1)(x1)+[f(2)−2f(1)](x2)=αx+βx2whereα,β∈Q.Converselyiff(x)=αx+βx2withα,β∈Qthenitisroutinelyverifiedthatfsatisfiestheconstraint.
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]
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
The possible functions are exactly those of the form f(x)=αx2+βx with α,β∈Q.
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,(z−3)a) for some z∈Z and a∈Q gives:
f(za)=13(2f(3a)−3f(2a))+2f((z−1)a)−f((z−2)a. [Eq.1]
Now we keep a≠0 fixed. We define dz:=f(za)−f((z−1)a) and C:=23f(3a)−f(2a) and rewrite [Eq.1] as:
dz=dz−1+C. 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].
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 z∈Z.
Since a was arbitrary, for a given r∈N there exist α1 and β1, such that:
f(zar)=α1z2+β1z.
Now pick an s∈Z 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 z∈Z and r∈N the number azr goes through all rationals. Therefore:
f(q)=αa2q2+βaq for all q∈Q, as desired.