Problems

Solution:

1. Claim: If \(n \in \mathbb{Z}\) then \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\) Proof: Since \(\lfloor x \rfloor \leq x\) then \(\lfloor x \rfloor + n \leq x + n\) and \(\lfloor x \rfloor + n \in \mathbb{Z}\) we must have that \(\lfloor x \rfloor +n \leq \lfloor x + n \rfloor\). However, since \(\lfloor x \rfloor + 1 > x\) we must also have that \(\lfloor x \rfloor + 1 + n > x + n\), therefore \(\lfloor x \rfloor + n\) is the largest integer less than \(x + n\) as required.
2. 1. Claim: \(f_n\left(x + \frac{1}{n}\right) = f_n(x)\) Proof: \begin{align*} f_n\left(x + \frac{1}{n}\right) &=\left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{1}{n}+ \frac{1}{n} \right\rfloor + \left\lfloor x+ \frac{1}{n} + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{1}{n} + \frac{n-1}{n} \right\rfloor - \left \lfloor n\left ( x + \frac{1}{n} \right) \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{n}{n} \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ 1 \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \lfloor x \rfloor + 1 - \left ( \lfloor nx \rfloor + 1 \right) \\ &= \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n-1}{n} \right\rfloor - \lfloor nx \rfloor \\ &= f_n(x) \end{align*}
   2. Suppose \(0 \leq t < \frac1n\), then note that \(\left \lfloor t + \frac{k}{n} \right \rfloor = 0\) for \(0 \leq k \leq n - 1\) and \(\lfloor n t \rfloor = 0\) since \(nt < 1\)
   3. Since \(f_n(x)\) is zero on \([0, \tfrac1n)\) and periodic with period \(\tfrac1n\) it must be constantly zero
3. 1. Claim: \(\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor = \lfloor x \rfloor\) Proof: Suppose \(x = n + \epsilon\) where \(0 \leq \epsilon < 1\), ie \(n = \lfloor x \rfloor\), then consider two cases: Case 1: \(n = 2k\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k + \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k + \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon}{2} \right\rfloor + k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor \\ &= 2k \\ &= n \end{align*} Case 2: \(n = 2k + 1\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k +1+ \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k +1+ \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor + k +1+ \left\lfloor \frac{\epsilon}{2} \right\rfloor \\ &= 2k +1\\ &= n \end{align*} as required.
   2. Since \(\left \lfloor \frac{x+1}{2} \right \rfloor = \lfloor x \rfloor - \lfloor \frac{x}{2} \rfloor\) and in general, \(\left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor = \lfloor \frac{x}{2^k} \rfloor - \lfloor \frac{x}{2^{k+1}} \rfloor\) and so in general: \begin{align*} \sum_{k=0}^\infty \left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor &= \sum_{k=0}^\infty \left ( \left \lfloor \frac{x}{2^k} \right \rfloor -\left \lfloor \frac{x}{2^{k+1}} \right \rfloor \right) \\ &= \lfloor x \rfloor \end{align*}

Solution:

1. \(\,\) \begin{align*} && f(0+x) &= f(0)f(x) \\ \Rightarrow && f(0) &= 0, 1\\ &&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\ \Rightarrow && f(0) &= 1 \end{align*} \begin{align*} && f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\ &&&= f(x) \cdot f'(0) \\ &&&= kf(x) \end{align*} Since \(f'(x) = kf(x)\) we must have \(\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}\) but \(f(0) = 1\) so \(f(x) = e^{kx}\)
2. Consider \begin{align*} && g(0+0) &= \frac{g(0)+g(0)}{1+(g(0))^2} \\ \Rightarrow && g(0)(1+(g(0))^2)&= 2g(0) \\ \Rightarrow && 0 &= g(0)\left (1- (g(0))^2 \right) \\ \Rightarrow && g(0) &= -1, 0, 1 \\ \Rightarrow && g(0) &= 0 \tag{\(|g(0)| < 1\)} \end{align*} \begin{align*} && g'(x) &=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)}-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{g(x)+g(h)-g(x)(1+g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= \lim_{h\to 0} \frac{g(h)-g(x)(g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= (1-(g(x))^2) \cdot \lim_{h \to 0} \frac{1}{1+g(x)g(h)} \cdot \lim_{h \to 0} \frac{g(h)}{h} \\ &&&= (1-(g(x))^2) \cdot \frac{1}{1+g(x)\cdot 0} \cdot \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ &&&= (1-(g(x))^2) \cdot g'(0)\\ &&&= k (1-(g(x))^2) \\ \end{align*} Let \(y = g(x)\) so \begin{align*} && y' &= k(1-y^2) \\ \Rightarrow && kx &= \int \frac{1}{1-y^2} \d y \\ \Rightarrow &&&= \int \frac12\left ( \frac{1}{1-y} + \frac{1}{1+y} \right) \d y \\ &&&= \frac12\ln \left ( \frac{1+y}{1-y} \right) + C \\ x = 0, y = 0: && 0 &= \ln 1 + C \\ \Rightarrow && C &= 0 \\ \Rightarrow && \frac{1+y}{1-y} &= e^{2kx} \\ \Rightarrow && 1+y &= e^{2kx} - e^{2kx}y \\ \Rightarrow && y &= \frac{e^{2kx}-1}{e^{2kx}+1} \\ &&&= \tanh kx \end{align*}

Solution:

1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

Solution:

1. \(\,\) \begin{align*} && T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\ &&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\ u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\ &&&= T_\infty - T(x^{-1}) \end{align*}
2. Let \(v = \frac{u+a}{1-au}\) then \begin{align*} && \frac{\d v}{\d u} &= \frac{(1-au) \cdot 1 - (u+a)\cdot(-a)}{(1-au)^2} \\ &&&= \frac{1-au+au+a^2}{(1-au)^2} \\ &&&= \frac{1+a^2}{(1-au)^2} \\ \\ && \frac{1+v^2}{1+u^2} &= \frac{1 + \left ( \frac{u+a}{1-ua} \right)^2}{1+u^2} \\ &&&= \frac{(1-ua)^2+(u+a)^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+u^2a^2+u^2+a^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{(1+u^2)(1+a^2)}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+a^2}{(1-ua)^2} \end{align*} if \(a > 0, x < \frac1a\) then \begin{align*} && T(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_{v=a}^{v=\frac{a+x}{1-ax}} \frac{1}{1+u^2} \frac{1+u^2}{1+v^2} \d v \\ &&&= T\left ( \frac{x+a}{1-ax} \right) - T(a) \\ \\ \Rightarrow && T(x^{-1}) &= T_\infty - T(x) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T(a) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T_\infty-T(a^{-1}) \\ &&&= 2T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) -T(a^{-1}) \end{align*} \(b > 0, y > \frac1b\) then \(y> 0, b > \frac1y\) (same as letting \(x = \frac1y, a = \frac1b\) \begin{align*} && T(y) &= 2T_\infty - 2T \left ( \frac{\frac1y+\frac1b}{1-\frac1{by}} \right) + T(b) \\ \Rightarrow && T(y) &= 2T_\infty - 2T \left ( \frac{b+y}{by-1} \right) + T(b) \\ \end{align*}
3. Letting \(y = b = \sqrt{3}\) in the final equation \begin{align*} && T(\sqrt{3}) &= 2T_{\infty} - T \left ( \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}\sqrt{3}-1} \right) -T (\sqrt{3}) \\ &&&= 2T_\infty - 2T(\sqrt{3}) \\ \Rightarrow && T(\sqrt{3}) &= \tfrac23 T_\infty \end{align*} Let \(x = \sqrt2 - 1, a = 1\) so, \begin{align*} && T(\sqrt2 -1) &= T \left ( \frac{\sqrt2-1+1}{1-\sqrt2+1} \right)-T(1) \\ &&&= T \left ( \frac{\sqrt{2}}{2-\sqrt{2}} \right) - T(1) \\ &&&= T(\frac{\sqrt{2}(2+\sqrt{2})}{2}) - T(1) \\ &&&= T(\sqrt{2}+1) - T(1) \\ &&&= T_\infty - T(\sqrt2-1)-T(1) \\ \Rightarrow && T(\sqrt{2}-1) &= \frac12T_\infty-\frac12T(1) \\ && T(1) &= T_\infty - T(1) \\ \Rightarrow && T(1) &= \frac12 T_\infty \\ \Rightarrow && T(\sqrt2-1) &= \frac12T_\infty - \frac14T_\infty \\ &&&= \frac14 T_\infty \end{align*}

Solution:

1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.

Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

Solution: \begin{align*} &&2f(x) &\equiv f(x) + f(x) \\ &&&\equiv f(x+x) \\ &&&\equiv f(2x) \\ \\ \Rightarrow && 2f(0) &= f(0) \\ \Rightarrow && f(0) &= 0 \\ && f''(0) &= \lim_{h \to 0} \frac{f(2h)-2f(0)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(2h)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(0)}{h^2} \\ &&&= 0 \\ \Rightarrow && f''(0) &= 0 \end{align*} If \(f(x)\) satisfies the equation, then \(f'(x)\) satisfies the equation. In particular this means that \(f^{(n)}(0) = 0\) for all \(n \geq 2\). Therefore the only non-zero term in the Maclaurin series is \(x^1\). Therefore \(f(x) = cx\)

1. Suppose \(g(x)g(y) \equiv g(x+y)\), then if \(G(x) = \ln g(x)\) we must have \(G(x)+G(y) \equiv G(x+y)\), ie \(G(x) = cx \Rightarrow g(x) = e^{cx}\)
2. Suppose \(h(x)+h(y) \equiv h(xy)\), then if \(h(e^u) = H(u)\) we must have that \(H(u)+H(v) \equiv h(e^u) + h(e^v) \equiv h(e^{u+v}) \equiv H(u+v)\).Therefore \(H(u) = cu\), ie \(h(e^u) = cu\) or \(h(x) = h(e^{\ln x}) = c \ln x\).
3. Finally if \(t(x) + t(y) \equiv t(z)\), the considering \(T(w) = t(\tan w)\) then \(T(x) + T(y) \equiv t(\tan x) + t(\tan y) \equiv t( \frac{\tan x + \tan y}{1- \tan x \tan y}) \equiv t (\tan (x+y)) \equiv T(x+y)\). Therefore \(T(x) = cx\) Therefore \(t(\tan w) = c w \Rightarrow t(x) = c \tan^{-1} x\)

Solution: \begin{align*} && g(0) &= f(0)+f(-0) = 2f(0) = 2 \\ && g'(x) &= f'(x) - f'(-x) \\ && g'(0) &= f'(0) - f'(-0) = 0 \\ && g''(x) &= f''(x) +f''(-x) \\ \Rightarrow && g''(x) + g(x) &= f''(x) +f''(-x) + f(x) + f(-x) \\ &&&= f''(x)+ f(-x) +f''(-x) + f(x) \\ &&&= x + 3 \cos 2x + (-x + 3 \cos (-2x) ) \\ &&&= 6 \cos 2x \\ \end{align*} Considering the homogeneous part, we should expected a solution of the form \(g(x) = A \sin x + B \cos x\). Seeking an integrating factor of the form \(g(x) = C \cos 2x\) we see that \(-4C \cos 2x + C \cos 2x = 6 \cos 2x \Rightarrow -3C = 6 \Rightarrow C = -2\). Therefore the general solution is \begin{align*} && g(x) &= A\sin x + B \cos x - 2\cos 2x \\ && g(0) &= B - 2 = 2\\ && g'(0) &= A = 0 \\ \Rightarrow && g(x) &= 4\cos x - 2\cos 2x \\ \end{align*} \begin{align*} && h(0) &= f(0) - f(-0) = 0 \\ && h'(x) &= f'(x) + f'(-x) \\ && h'(0) &= f'(0) + f'(-0) = -2 \\ && h''(x) &= f''(x) - f''(-x) \\ \Rightarrow && h''(x) - h(x) &= f''(x) - f''(-x) -( f(x) - f(-x)) \\ &&&= f''(x) +f(-x)- (f''(-x) + f(x)) \\ &&&= x + 3\cos 2x - (-x + 3 \cos(-2x)) \\ &&&= 2x \end{align*} Considering the homogeneous part, we should expect a solution of the form \(Ae^x + Be^{-x}\). For a specific integral, we can take \(-2x\), ie \begin{align*} && h(x) &= Ae^x + Be^{-x} - 2x \\ && h(0) &= A+B =0 \\ && h'(0) &= A-B-2 =-2 \\ \Rightarrow && A &=B = 0 \\ \Rightarrow && h(x) &= -2x \end{align*} Therefore \(f(x) = \frac12(f(x) + f(-x)) + \frac12(f(x) -f(-x)) = 2\cos x - \cos 2x -x\)

Solution: Let \(P(x,y)\) be the statement that the functional equation holds, then: \begin{align*} P(0,0): && f(0) &= f(0)f(0)-f(a)f(a) \\ \Rightarrow && 1 &= 1 - f(a)^2 \\ \Rightarrow && f(a)^2 &= 0 \\ \Rightarrow && f(a) &= 0 \end{align*}

1. \begin{align*} P(0,t): && f(-t) &= f(0)f(t) - f(a)f(a-t) \\ \Rightarrow && f(-t) &= f(t) - 0 \\ \Rightarrow && f(t) &= f(-t) \end{align*}
2. \begin{align*} P(a,a): && f(0) &= f(a)f(a)-f(0)f(2a) \\ \Rightarrow && 1 &= 0 - f(2a) \\ \Rightarrow && f(2a) &= -1 \end{align*}
3. \begin{align*} P(2a,t): && f(2a-t) &= f(2a)f(t) - f(-a)f(a+t) \\ \Rightarrow && f(2a-t) &= -f(t)-f(a)f(a+t) \\ &&&= -f(t)-0 \\ \Rightarrow && f(2a-t) &= -f(t) \end{align*}
4. \begin{align*} && f(4a+t) &= f(2a-(-2a-t)) \\ &&&=-f(2a+t) \\ &&&=-f(2a-(-t)) \\ &&&=f(-t) \\ &&&=f(t) \end{align*}

Solution: Note that \(z = xy/(x+y+1) \Rightarrow y(x-z) = z(x+1)\) \begin{align*} && g(x) + g(y) &= g(z) \\ \Rightarrow && g'(x) &= g'(z) \cdot \frac{y(x+y+1) - xy \cdot 1} {(x+y+1)^2} \\ &&&= g'(z) \frac{y^2+y}{(x+y+1)^2} \\ &&&= g'(z) \frac{z^2(y^2+y)}{x^2y^2} \\ &&&= g'(z) \frac{z^2(y+1)}{x^2y} \\ &&&= g'(z) \frac{z^2}{x^2} \left (1 + \frac{x-z}{z(x+1)} \right) \\ &&&= g'(z) \frac{z}{x^2} \frac{zx+x}{x+1} \\ &&&= g'(z) \frac{z(z+1)}{x(x+1)} \end{align*} If \(x = 1\) then as \(y\) ranges from \(0\) to \(\infty\), \(z\) ranges from \(0\) to \(1\), so \(g'(1) = \frac{z(z+1)}{1(1+1)}g'(z)\), ie \(2g'(1) = (u^2+u)g'(u)\). \begin{align*} && g'(u) &= \frac{A}{u(u+1)} \\ \Rightarrow && g(u) &= A\int \left ( \frac{1}{u} - \frac{1}{u+1} \right) \d u \\ &&&= A \left ( \ln u - \ln(u+1) \right) + B \\ &&&= A \ln \left ( \frac{u}{u+1} \right) + B \\ \\ && A \ln \left ( \frac{x}{x+1} \right) + B+A \ln \left ( \frac{y}{y+1} \right) + B &=A \ln \left ( \frac{z}{z+1} \right) + B \\ \Rightarrow && B &= A \ln \left ( \frac{z}{z+1} \frac{y+1}{y} \frac{x+1}{x} \right) \\ &&&= A \ln \left ( \frac{1}{1+\frac{x+y+1}{xy}} \frac{(y+1)(x+1)}{xy} \right) \\ &&&= A \ln 1 \\ &&& = 0 \end{align*} Therefore \(B = 0\). \(A\) cannot be determined from \((*)\). Suppose \(f(x) + f(y) = f(z)\), then \(f'(x) = yf'(z)\). Letting \(x = 1\) we find \(f'(1) = uf'(u) \Rightarrow f(u) = C \ln u + D\), but \(D = 0\) so \(f(x) = C \ln x\)