Functional Equations: Problems with Solutions

Solution:
We find the value of [tex]x[/tex] for which the argument [tex]5x-4[/tex] equals [tex]6[/tex]:
[tex]5x-4=6\Rightarrow x=2[/tex]
Substituting [tex]x=2[/tex] into the given equality, we get
[tex]f(6)=f(5\times 2-4)=2\times 2+1=5[/tex]

Solution:
Putting [tex]x=1[/tex] and [tex]x=0[/tex] into the condition, we get the system
[tex]\left\{\begin{array}{l} 2f(1)+f(0)=1 \\ f(1)+2f(0)=-1 \end{array}\right.[/tex]
From the first equation [tex]f(0)=1-2f(1)[/tex]. Substituting into the second one:
[tex]f(1)+2-4f(1)=-1\Rightarrow 3f(1)=3\Rightarrow f(1)=1[/tex]
(Such a function exists: [tex]f(x)=2x-1[/tex] satisfies the condition, since [tex]2(2x-1)+\left(2(1-x)-1\right)=2x-1[/tex].)

Solution:
For each function we compute the sum [tex]f(x)+f(-x)[/tex]:
[tex]f(x)=\frac{x^2}{2}:\quad \frac{x^2}{2}+\frac{x^2}{2}=x^2[/tex]
[tex]f(x)=\frac{x^2+x}{2}:\quad \frac{x^2+x}{2}+\frac{x^2-x}{2}=x^2[/tex]
[tex]f(x)=\frac{x^2}{2}+5x-x^3:\quad \left(\frac{x^2}{2}+5x-x^3\right)+\left(\frac{x^2}{2}-5x+x^3\right)=x^2[/tex]
[tex]f(x)=\frac{x^2+5x+1}{2}:\quad \frac{x^2+5x+1}{2}+\frac{x^2-5x+1}{2}=x^2+1\ne x^2[/tex]
So the first three functions are of class [tex]2[/tex], and only [tex]f(x)=\frac{x^2+5x+1}{2}[/tex] is not (for example, at [tex]x=0[/tex] the sum equals [tex]1\ne 0[/tex]).

Solution:
Let [tex]P(x)=ax^2+bx+c[/tex], where [tex]a\ne 0[/tex]. From [tex]P(0)=5[/tex] we get [tex]c=5[/tex]. Further,
[tex]P(x+1)-P(x)=a(x+1)^2+b(x+1)+c-ax^2-bx-c=2ax+a+b[/tex]
This must equal [tex]8x+3[/tex] for every [tex]x[/tex], so the coefficients of equal powers of [tex]x[/tex] coincide:
[tex]\left\{\begin{array}{l} 2a=8 \\ a+b=3 \end{array}\right.\Rightarrow a=4, b=-1[/tex]
Hence [tex]P(x)=4x^2-x+5[/tex]. Check:
[tex]4(x+1)^2-(x+1)+5-\left(4x^2-x+5\right)=8x+4-1=8x+3[/tex]

Solution:
Choose [tex]x[/tex] and [tex]y[/tex] so that the two unknown values [tex]f(2x+y)[/tex] and [tex]f(3x-y)[/tex] become equal: [tex]2x+y=3x-y[/tex], i.e. [tex]x=2y[/tex]. Since we need [tex]f(10)[/tex], take [tex]x=10[/tex], [tex]y=5[/tex]; then [tex]2x+y=3x-y=25[/tex] and the condition becomes
[tex]f(10)+f(25)+5\times 10\times 5=f(25)+2\times 10^2+1[/tex]
[tex]f(10)+250=201\Rightarrow f(10)=-49[/tex]
(Such a function exists: for [tex]x=2y[/tex] the condition gives [tex]f(2y)+10y^2=8y^2+1[/tex], so [tex]f(t)=1-\frac{t^2}{2}[/tex], and a direct check shows that this function satisfies the condition for all [tex]x[/tex] and [tex]y[/tex].)

Solution:
Put [tex]a=4[/tex], [tex]b=1[/tex]. Since [tex]\frac{4+2}{3}=2[/tex], we get
[tex]f(2)=\frac{f(4)+2f(1)}{3}=\frac{8+4}{3}=4[/tex]
Put [tex]a=1[/tex], [tex]b=4[/tex]. Since [tex]\frac{1+8}{3}=3[/tex], we get
[tex]f(3)=\frac{f(1)+2f(4)}{3}=\frac{2+16}{3}=6[/tex]
Now put [tex]a=n[/tex], [tex]b=n+3[/tex]. Since [tex]\frac{n+2(n+3)}{3}=n+2[/tex], we get [tex]f(n+2)=\frac{f(n)+2f(n+3)}{3}[/tex], i.e.
[tex]f(n+3)=\frac{3f(n+2)-f(n)}{2}[/tex]
We prove by induction that [tex]f(n)=2n[/tex] for every natural [tex]n[/tex]. It holds for [tex]n=1, 2, 3[/tex]. If [tex]f(m)=2m[/tex] for all [tex]m\le n+2[/tex], then
[tex]f(n+3)=\frac{3\times 2(n+2)-2n}{2}=\frac{4n+12}{2}=2(n+3)[/tex]
Therefore
[tex]f(1999)=2\times 1999=3998[/tex]
(The function [tex]f(x)=2x[/tex] satisfies all the conditions, so the data are consistent.)

Solution:
Since [tex]x\ne 0[/tex] implies [tex]\frac{1}{x}\ne 0[/tex], we may replace [tex]x[/tex] by [tex]\frac{1}{x}[/tex]:
[tex]f\left(\frac{1}{x}\right)+3f(x)=\frac{1}{x^2}[/tex]
Multiplying this equality by [tex]3[/tex] and subtracting the original one, we get
[tex]8f(x)=\frac{3}{x^2}-x^2\Rightarrow f(x)=\frac{1}{8}\left(\frac{3}{x^2}-x^2\right)[/tex]
Check: [tex]f(x)+3f\left(\frac{1}{x}\right)=\frac{1}{8}\left(\frac{3}{x^2}-x^2+9x^2-\frac{3}{x^2}\right)=x^2[/tex].
Now the condition on [tex]u[/tex]. Since [tex]\log_2 t\ge 0\Leftrightarrow t\ge 1[/tex] and the base [tex]\frac{1}{2}[/tex] is less than [tex]1[/tex], the inequality is equivalent to
[tex]\log_{\frac{1}{2}}\frac{u-1}{u+1}\ge 1\Leftrightarrow 0\lt\frac{u-1}{u+1}\le\frac{1}{2}[/tex]
The left inequality gives [tex]u\lt -1[/tex] or [tex]u\gt 1[/tex]; the right one gives
[tex]\frac{u-1}{u+1}-\frac{1}{2}=\frac{u-3}{2(u+1)}\le 0\Leftrightarrow -1\lt u\le 3[/tex]
Together, [tex]u\in\left(1, 3\right][/tex]. On this interval both [tex]\frac{3}{u^2}[/tex] and [tex]-u^2[/tex] are decreasing, so [tex]f(u)[/tex] is decreasing and attains its least value at [tex]u=3[/tex]:
[tex]f(3)=\frac{1}{8}\left(\frac{1}{3}-9\right)=-\frac{13}{12}[/tex]
(The value [tex]f(1)=\frac{1}{4}[/tex] is not attained, because [tex]u=1[/tex] does not satisfy the condition.)

Solution:
The idea is to choose [tex]y[/tex] so that [tex]xy=x+y[/tex]; then [tex]f(xy)[/tex] and [tex]f(x+y)[/tex] cancel. For [tex]x\ne 1[/tex] put [tex]y=\frac{x}{x-1}[/tex]. Then
[tex]xy=\frac{x^2}{x-1},\quad x+y=\frac{x(x-1)+x}{x-1}=\frac{x^2}{x-1}[/tex]
so [tex]xy=x+y[/tex] and the given inequality becomes [tex]f(x-y)\ge 0[/tex], where
[tex]x-y=\frac{x(x-1)-x}{x-1}=\frac{x^2-2x}{x-1}[/tex]
It remains to show that every real number [tex]a[/tex] can be written in this form for some [tex]x\ne 1[/tex]. For [tex]x\ne 1[/tex]
[tex]\frac{x^2-2x}{x-1}=a\Leftrightarrow x^2-(a+2)x+a=0[/tex]
The discriminant of this quadratic equation is [tex](a+2)^2-4a=a^2+4\gt 0[/tex], so it has real roots, and [tex]x=1[/tex] is not a root, because [tex]1-(a+2)+a=-1\ne 0[/tex].
Taking such a root [tex]x[/tex] and [tex]y=\frac{x}{x-1}[/tex], we get [tex]x-y=a[/tex] and hence [tex]f(a)\ge 0[/tex]. Since [tex]a[/tex] is an arbitrary real number, [tex]f(x)\ge 0[/tex] for every real [tex]x[/tex].

Solution:
Use the identities [tex]3xy(x+y)=(x+y)^3-x^3-y^3[/tex] and [tex]18xy=9\left((x+y)^2-x^2-y^2\right)[/tex]. They give
[tex]3xy(x+y+6)=\left[(x+y)^3+9(x+y)^2\right]-\left[x^3+9x^2\right]-\left[y^3+9y^2\right][/tex]
Let [tex]h(x)=f(x)-x^3-9x^2-8[/tex]. Substituting [tex]f(x)=h(x)+x^3+9x^2+8[/tex] into the equation, the cubic and quadratic terms cancel and the constants give [tex]8=8+8-8[/tex]. We obtain Cauchy's equation
[tex]h(x+y)=h(x)+h(y),\quad x,y\in\mathbb{Q}[/tex]
With [tex]x=y=0[/tex] we get [tex]h(0)=0[/tex]. By induction [tex]h(nx)=nh(x)[/tex] for every natural [tex]n[/tex], since [tex]h((n+1)x)=h(nx)+h(x)[/tex]. With [tex]y=-x[/tex] we get [tex]h(-x)=-h(x)[/tex], so [tex]h(nx)=nh(x)[/tex] for every integer [tex]n[/tex]. For [tex]x=\frac{p}{q}[/tex] with [tex]p\in\mathbb{Z}[/tex], [tex]q\in\mathbb{N}[/tex]:
[tex]q\,h\left(\frac{p}{q}\right)=h(p)=p\,h(1)\Rightarrow h\left(\frac{p}{q}\right)=\frac{p}{q}\,h(1)[/tex]
Hence [tex]h(x)=cx[/tex] with [tex]c=h(1)[/tex], i.e. [tex]f(x)=x^3+9x^2+cx+8[/tex]. Conversely, every such function satisfies the equation, because [tex]h(x)=cx[/tex] is additive. The condition [tex]f(-8)=0[/tex] gives
[tex]-512+576-8c+8=0\Rightarrow c=9[/tex]
Therefore the only such function is [tex]f(x)=x^3+9x^2+9x+8=(x+8)\left(x^2+x+1\right)[/tex].

Solution:
For [tex]n=1[/tex]: [tex]2P(1)=P(1)[/tex], so [tex]P(1)=0[/tex]. Subtracting the equality for [tex]n-1[/tex] from the equality for [tex]n[/tex] ([tex]n\ge 2[/tex]) gives [tex]P(2n-1)=2nP(n)-2(n-1)P(n-1)[/tex]. Both sides are polynomials in [tex]n[/tex] that agree for infinitely many values, so identically
[tex]P(2x-1)=2xP(x)-2(x-1)P(x-1)[/tex]
[tex]P[/tex] is not the zero polynomial, since [tex]P(2)=6[/tex]. Let [tex]d=\deg P[/tex] and let [tex]a\ne 0[/tex] be its leading coefficient. The leading term of the left side is [tex]2^da\,x^d[/tex]. With [tex]Q(x)=xP(x)[/tex] the right side equals [tex]2\left(Q(x)-Q(x-1)\right)[/tex], and since [tex]x^{d+1}-(x-1)^{d+1}=(d+1)x^d+\ldots[/tex], its leading term is [tex]2a(d+1)x^d[/tex]. Hence
[tex]2^d=2(d+1)[/tex]
This fails for [tex]d=0, 1, 2[/tex] ([tex]1\ne 2[/tex], [tex]2\ne 4[/tex], [tex]4\ne 6[/tex]) and holds for [tex]d=3[/tex]; for [tex]d\ge 4[/tex] we have [tex]2^d\gt 2(d+1)[/tex] (true for [tex]d=4[/tex], and when [tex]d[/tex] increases by [tex]1[/tex], the left side doubles while the right side grows only by [tex]2[/tex]). So [tex]\deg P=3[/tex].
Further values: [tex]n=2[/tex] gives [tex]4P(2)=P(1)+P(3)[/tex], so [tex]P(3)=24[/tex]; [tex]n=3[/tex] gives [tex]6P(3)=P(1)+P(3)+P(5)[/tex], so [tex]P(5)=144-24=120[/tex].
The cubic polynomial [tex]x^3-x[/tex] takes the values [tex]0, 6, 24, 120[/tex] at [tex]x=1, 2, 3, 5[/tex]. Two polynomials of degree at most [tex]3[/tex] that agree at [tex]4[/tex] points coincide (their difference has degree at most [tex]3[/tex] and [tex]4[/tex] roots), so
[tex]P(x)=x^3-x=(x-1)x(x+1)[/tex]
Check: [tex]\sum_{k=1}^{n}(2k-1)^3=n^2\left(2n^2-1\right)[/tex] and [tex]\sum_{k=1}^{n}(2k-1)=n^2[/tex], hence
[tex]P(1)+P(3)+\ldots+P(2n-1)=2n^4-2n^2=2n\left(n^3-n\right)=2nP(n)[/tex]
Therefore
[tex]P(2009)=2008\times 2009\times 2010=8108484720[/tex]

Solution:
The function [tex]f(n)=n[/tex] satisfies the condition: [tex]3n-2n=n[/tex].
Conversely, let [tex]f[/tex] satisfy the condition and put [tex]g(n)=f(n)-n[/tex]; [tex]g[/tex] takes integer values. The condition can be written as [tex]2f(f(n))-2f(n)=f(n)-n[/tex], i.e.
[tex]2g\left(f(n)\right)=g(n),\quad n\in\mathbb{Z}[/tex]
Fix an integer [tex]n[/tex] and define [tex]n_0=n[/tex], [tex]n_{k+1}=f(n_k)[/tex]; all [tex]n_k[/tex] are integers. Applying the relation to [tex]n_k[/tex] gives [tex]g(n_{k+1})=\frac{g(n_k)}{2}[/tex], so by induction, for every natural [tex]k[/tex],
[tex]g(n_k)=\frac{g(n)}{2^k}[/tex]
Since [tex]g(n_k)[/tex] is an integer, [tex]g(n)[/tex] is divisible by [tex]2^k[/tex] for every natural [tex]k[/tex]. If [tex]g(n)\ne 0[/tex], choose [tex]k[/tex] with [tex]2^k\gt\left|g(n)\right|[/tex]; a non-zero integer cannot be divisible by a number greater than its absolute value, a contradiction. Hence [tex]g(n)=0[/tex], i.e. [tex]f(n)=n[/tex] for every integer [tex]n[/tex].

Solution:
Suppose such a function exists and let [tex]g(x)=x^2-x-1[/tex], so [tex]f(f(x))=g(x)[/tex].
Step [tex]1[/tex]. [tex]f[/tex] commutes with [tex]g[/tex]: for every [tex]x[/tex]
[tex]f(g(x))=f\left(f(f(x))\right)=g\left(f(x)\right)[/tex]
Step [tex]2[/tex]. The fixed points of [tex]g[/tex]: [tex]x^2-x-1=x\Leftrightarrow x^2-2x-1=0[/tex], so they form the set [tex]A=\left\{1-\sqrt{2}, 1+\sqrt{2}\right\}[/tex]. For the equation [tex]g(g(x))=x[/tex] we have
[tex]g(g(x))-x=\left(x^2-x-1\right)^2-\left(x^2-x-1\right)-1-x[/tex]
[tex]=x^4-2x^3-2x^2+2x+1=\left(x^2-2x-1\right)\left(x^2-1\right)[/tex]
so its solutions form the set [tex]B=\left\{1-\sqrt{2}, 1+\sqrt{2}, -1, 1\right\}[/tex]. Note that [tex]g(1)=-1[/tex] and [tex]g(-1)=1[/tex].
Step [tex]3[/tex]. If [tex]g(p)=p[/tex], then by Step [tex]1[/tex] we get [tex]g(f(p))=f(g(p))=f(p)[/tex], so [tex]f[/tex] maps [tex]A[/tex] into [tex]A[/tex]. If [tex]g(g(t))=t[/tex], then [tex]g\left(g(f(t))\right)=f\left(g(g(t))\right)=f(t)[/tex], so [tex]f[/tex] maps [tex]B[/tex] into [tex]B[/tex].
Step [tex]4[/tex]. By Step [tex]3[/tex], [tex]f(1)\in B[/tex]. If [tex]f(1)\in A[/tex], then also [tex]f(f(1))\in A[/tex]; but [tex]f(f(1))=g(1)=-1\notin A[/tex]. Hence [tex]f(1)\in\{1, -1\}[/tex]. If [tex]f(1)=1[/tex], then [tex]g(1)=f(f(1))=f(1)=1[/tex], which contradicts [tex]g(1)=-1[/tex]. So [tex]f(1)=-1[/tex].
Then [tex]f(-1)=f(f(1))=g(1)=-1[/tex], hence [tex]g(-1)=f(f(-1))=f(-1)=-1[/tex], which contradicts [tex]g(-1)=1[/tex]. Therefore no such function exists.