Evaluating a Function

With $f(1)$ given, we can calculate $f(n)$ for all $n\in\mathbb{N}, n\geq1$ using the $f(x+y)$ formula inductively, with $x=n-1$ and $y=1$ .
This gives $f(5)=120$ .

Similarly, we can express $f(nx)$ from only $x$ and $f(x)$ , substituting $y=(n-1)x$ into the addition formula. This gives a slightly complicated, but manageable equation for $f(3x)$ . Substituting $x=\frac53$ the only unknown term is $f(\frac53)$ , which then can be easily calculated. $f(\frac53)=\frac{280}{27}$ , so the solution is $280+27=307$ .

We start by dividing through by $(x+y+1)$ , noting that if $x + y = -1$ , then $f(-1) = 0$ .

This yields $\frac{f(x+y)}{x+y+1} = \frac{f(x)}{x+1} + \frac{f(y)}{y+1} + xy$ , so we can simplify this by forcing $f(x) = (x+1)g(x)$ to define the function $g$ . This means that $g(x+y) = g(x) + g(y) + xy$ .

The next move may seem magic, so here is a motivation; $x = y = 0$ gives $g(0) = 0$ , and $y = -x$ gives $x^2 = g(x) + g(-x)$ . So we know that $g(x)$ is $\frac{x^2}{2}$ plus an odd function $h(x)$ :

If $g(x) = h(x) + \frac{x^2}{2}$ , then our equation reduces to $h(x + y) = h(x) + h(y)$ .

This is the Cauchy equation , and its solution is well known; on the rationals, at least, we know that $h(x) = kx$ for some constant $k$ .

Returning to $f$ , we see that $f(x) = \frac{1}{2}x(x+1)(x+k)$ on the rationals. Substituting in $x = 1$ gives $k = 3$ , and substituting in $x = \frac{5}{3}$ gives an answer of $f(x) = \frac{280}{27}$ .

On first glance, one will notice that $\frac {5}{3} = 1+\frac {2}{3}$ . So, to find $f(\frac {5}{3})$ , the only unknown that needs to be found is $f(\frac {2}{3})$ , since $f(1) = 4$ is already given.

To start off, it can be seen that $f(1)$ can be rewritten in the form of $f(x+y)$ , where $x=\frac {1}{3}$ and $y=\frac {2}{3}$ .

Substituting the variables into the function and simplifying,

$f(\frac {1}{3} + \frac {2}{3}) = (1+ \frac {\frac{2}{3}}{\frac {1}{3}+1})f(\frac {1}{3})+ (1+ \frac {\frac{1}{3}}{\frac {2}{3}+1})f(\frac {2}{3})+(\frac {1}{3})^2(\frac {2}{3})+(\frac {1}{3})(\frac {2}{3})+(\frac {1}{3})(\frac {2}{3})^2 = 4$

${\frac {3}{2}}f(\frac {1}{3})+ {\frac {6}{5}}f(\frac {2}{3})+\frac {2}{27}+\frac {2}{9}+\frac {4}{27} = 4$

At the end, equation 1 is obtained: ${\frac {3}{2}}f(\frac {1}{3})+ {\frac {6}{5}}f(\frac {2}{3}) = \frac {32}{9}$

At this point, we need another equation to solve for the value of $f(\frac {2}{3})$ . This is conveniently provided by setting $x+y=\frac {2}{3}$ , where $x=y=\frac {1}{3}$ .

Substituting these values into the function and simplying, we obtain:

$f(\frac {1}{3} + \frac {1}{3}) = (1+ \frac {\frac{1}{3}}{\frac {1}{3}+1})f(\frac {1}{3})+ (1+ \frac {\frac{1}{3}}{\frac {1}{3}+1})f(\frac {1}{3})+(\frac {1}{3})^2(\frac {1}{3})+(\frac {1}{3})(\frac {1}{3})+(\frac {1}{3})(\frac {1}{3})^2$

$f(\frac {2}{3}) = {\frac{5}{4}}f(\frac {1}{3})+ {\frac{5}{4}}f(\frac {1}{3})+\frac {1}{27}+\frac {1}{9}+\frac {1}{27}$

Equation 2 is then obtained: $f(\frac {2}{3})= {\frac{5}{2}}f(\frac {1}{3})+\frac {5}{27}$

By substituting equation 2 into equation 1, we can find the value of $f(\frac {1}{3})$ :

${\frac {3}{2}}f(\frac {1}{3})+ {\frac {6}{5}}({\frac{5}{2}}f(\frac {1}{3})+\frac {5}{27})=\frac {32}{9}$

${\frac {3}{2}}f(\frac {1}{3})+ {3}f(\frac {1}{3})+\frac {2}{9} = \frac {32}{9}$

${\frac {9}{2}}f(\frac {1}{3})=\frac {10}{3}$

$\Rightarrow f(\frac {1}{3}) = {\frac {20}{27}}$

The value of $f(\frac {1}{3})$ can then be substituted into equation 2 to find the value of $f(\frac {2}{3})$ :

$f(\frac {2}{3})= {\frac {5}{2}}(\frac {20}{27})+\frac {5}{27}$

$=\frac {50}{27} + \frac {5}{27} = \frac {55}{27}$

Finally, simply substitute the values $x=1$ , $y=\frac {2}{3}$ , $f(1)=4$ and $f(\frac {2}{3})=\frac {55}{27}$ into the function to find the value of $f(\frac {5}{3})$ :

$f(1 + \frac {2}{3}) = (1+ \frac {\frac{2}{3}}{{1+1}}){f(1)}+ (1+ \frac {1}{\frac {2}{3}+1}){f(\frac {2}{3})}+(1)^2(\frac {2}{3})+(1)(\frac {2}{3})+(1)(\frac {2}{3})^2$

$f(\frac {5}{3}) = {\frac {4}{3}}(4)+ {\frac {8}{5}}(\frac {55}{27})+\frac {2}{3}+\frac {2}{3}+\frac {4}{9}$

$=\frac {16}{3} + \frac {88}{27} + \frac {16}{9}$

$=\frac {280}{27}$

Thus, the answer is $280+27=307$ .

Substitute x = 1/3 and y = 2/3,

f(1) = (3/2) f(1/3) + (6/5) f(2/3) + 4/9

(6/5) f(2/3) = 32/9 - (3/2) f(1/3)

f(2/3) = 80/27 - (5/4)*f(1/3) .... (1)

Then substitute x = y = 1/3,

f(2/3) = (5/2)*f(1/3) + 5/27

(5/2)*f(1/3) = f(2/3) - 5/27

f(1/3) = (2/5)*f(2/3) - 2/27

Substitute this form to (1)

f(2/3) = 80/27 - (5/4)((2/5)*f(2/3) - 2/27)

f(2/3) = 165/54 - (1/2)*f(2/3)

(3/2)*f(2/3) = 165/54

f(2/3) = 55/27

Substitute x = 2/3 and y = 1,

f(5/3) = (8/5) f(2/3) + (4/3) f(1) + 16/9

f(5/3) = 88/27 + 16/3 + 16/9

f(5/3) = 280/27

If we substitute $x = 1$ and $y = 1$ into the given equation and simplify, we get:

$f(2) = 3f(1)+3$

Since $f(1) = 4$ , we know that $f(2) = 15$ .

Now let $x = \frac{1}{3}$ and $y = \frac{2}{3}$ . We get:

$f(1) = 4 = \frac{3}{2}f( \frac{1}{3} ) + \frac{6}{5}f( \frac{2}{3} ) + \frac{4}{9}$ ...........eqt (1)

Now let $x= \frac{5}{3}$ and $y = \frac{1}{3}$ . Now we get:

$f(2) = 15 = \frac{9}{8}f( \frac{5}{3} ) + \frac{9}{4}f( \frac{1}{3} ) + \frac{5}{3}$ ...........eqt(2)

Finally, let $x = \frac{1}{3}$ and $y = \frac{1}{3}$ . Substitution gives:

$f( \frac{2}{3} ) = \frac{5}{2}f( \frac{1}{3} ) + \frac{5}{27}$ ............eqt(3)

Equations (1), (2), and (3) constitute a system of three equations in the three unknowns $f( \frac{1}{3} )$ , $f( \frac{2}{3} )$ , and $f( \frac{5}{3} )$ . After some straightforward algebra, we obtain:

$f( \frac{1}{3} ) = \frac{20}{27} \\ f( \frac{2}{3} ) = \frac{55}{27} \\ f( \frac{5}{3} ) = \frac{280}{27}$

So our answer is $280 + 27 = 307$ .

$f(x + y) = (x + y + 1)[\frac{f(x)}{x + 1} + \frac{f(y)}{y + 1} + xy]$ .

Substituting $y = 1$ gives $f(x + 1) = (x + 2)[\frac{f(x)}{x + 1} + x + 2]$ . We claim that if $n \in \mathbb{N}$ , the. $f(n) = (n + 1)[\frac{(n + 2)(n + 3)}{2} - 1]$ . Base case is clearly established. Inductive step, we claim that $f(n + 1) = (n + 2)[\frac{(n + 2)(n + 3)}{2} - 1]$ .

$f(n + 1) = (n + 2)[\frac{f(n)}{n + 1} + n + 2] = (n + 2)\frac{(n + 4)(n + 1)}{2} = (n + 2)[\frac{(n + 2)(n + 3)}{2} - 1]$ . This establishes our claim.

Now, $f(2x) = (2x + 1)[\frac{2f(x)}{x + 1} + x^{2}], f(3x) = (3x + 1)[\frac{f(x)}{x + 1} + \frac{f(2x)}{2x + 1} + 2x^{2}]$ . Substituting $x = y = 0$ gives $f(0) = 0$ . Substituting $x = 1$ , $f(1) = 2[\frac{f(\frac{1}{3})}{\frac{4}{3}} + \frac{f(\frac{2}{3})}{\frac{5}{3}} + \frac{2}{9}]$ . Let $f(\frac{1}{3}) = a, f(\frac{2}{3}) = b$ . Then substituting $x = \frac{1}{3}$ in the $f(2x)$ equation gives $\frac{5}{2}a - b = -\frac{5}{27} , \frac{3}{4}a + \frac{3}{5}b = \frac{16}{9}$ . Solving for b gives $b = \frac{55}{27}$ . Substituting and solving for $f(\frac{5}{3})$ in the $f(n + 1)$ equation above gives $f(\frac{5}{3}) = \frac{280}{27}$ and the answer is $307$ .

Given that

$f(x+y) = ( 1 + \frac {y}{x+1}) f(x) + (1 + \frac {x}{y+1}) f(y) + x^2y + xy + xy^2$

$f(x+y) = ( \frac {x+1+y}{x+1}) f(x) + ( \frac {y+1+x}{y+1}) f(y) + x^2y + xy^2 + xy$

$f(x+y) = ( \frac {x+y+1}{x+1}) f(x) + ( \frac {x+y+1}{y+1}) f(y) + xy(x+y+1)$

$f(x+y) = (x+y+1)(\frac {f(x)}{x+1} + \frac {f(y)}{y+1} + xy)$

Since $f(1) = 4$ , we could manipulate the function so that

$f(\frac {1}{3} + \frac {2}{3}) = 4$

$(\frac {1}{3}+\frac {2}{3}+1)(\frac {f(\frac {1}{3})}{\frac {1}{3}+1} + \frac {f(\frac {2}{3})}{\frac {2}{3}+1} + (\frac {1}{3})(\frac {2}{3})) = 4$

$(2)(\frac {f(\frac {1}{3})}{\frac {4}{3}} + \frac {f(\frac {2}{3})}{\frac {5}{3}} + \frac {2}{9}) = 4$

$(2)(\frac {3}{4} f(\frac {1}{3}) + \frac {3}{5} f(\frac {2}{3}) + \frac {2}{9}) = 4$

$\frac {3}{4} f(\frac {1}{3}) + \frac {3}{5} f(\frac {2}{3}) + \frac {2}{9} = 2$

$\frac {3}{4} f(\frac {1}{3}) + \frac {3}{5} f(\frac {2}{3}) = 2 - \frac {2}{9}$

$\frac {3}{4} f(\frac {1}{3}) + \frac {3}{5} f(\frac {2}{3}) = \frac {16}{9} . . . 1)$

From another way, we also could manipulate the function so that $f(\frac {1}{3} + \frac {1}{3}) = (\frac {1}{3}+\frac {1}{3}+1)(\frac {f(\frac {1}{3})}{\frac {1}{3}+1} + \frac {f(\frac {1}{3})}{\frac {1}{3}+1} + (\frac {1}{3})(\frac {1}{3}))$

$f(\frac {2}{3}) = (\frac {5}{3})(\frac {f(\frac {1}{3})}{\frac {4}{3}} + \frac {f(\frac {1}{3})}{\frac {4}{3}} + \frac {1}{9})$

$f(\frac {2}{3}) = (\frac {5}{3})(\frac {3}{4} f(\frac {1}{3}) + \frac {3}{4} f(\frac {1}{3}) + \frac {1}{9})$

$(\frac {3}{5}) f(\frac {2}{3}) = \frac {3}{4} f(\frac {1}{3}) + \frac {3}{4} f(\frac {1}{3}) + \frac {1}{9}$

$(\frac {3}{5}) f(\frac {2}{3}) = \frac {3}{2} f(\frac {1}{3}) + \frac {1}{9}$

$(\frac {3}{5}) f(\frac {2}{3}) - \frac {1}{9} = \frac {3}{2} f(\frac {1}{3})$

$\frac {3}{2} f(\frac {1}{3}) = (\frac {3}{5}) f(\frac {2}{3}) - \frac {1}{9}$

$f(\frac {1}{3}) = (\frac {2}{3})(\frac {3}{5} f(\frac {2}{3}) - \frac {1}{9})$

$f(\frac {1}{3}) = \frac {2}{5} f(\frac {2}{3}) - \frac {2}{27} . . . 2)$

By combining equation 1) and 2), we got that $(\frac {3}{4})(\frac {2}{5} f(\frac {2}{3}) - \frac {2}{27}) + \frac {3}{5} f(\frac {2}{3}) = \frac {16}{9}$

$\frac {3}{10} f(\frac {2}{3}) - \frac {1}{18} + \frac {6}{10} f(\frac {2}{3}) = \frac {16}{9}$

$\frac {9}{10} f(\frac {2}{3}) = \frac {32}{18} + \frac {1}{18}$

$\frac {9}{10} f(\frac {2}{3}) = \frac {33}{18}$

$f(\frac {2}{3}) = (\frac {10}{9})(\frac {33}{18})$

$f(\frac {2}{3}) = \frac {55}{27} . . . 3)$

Now, we have $f(1) = 4$ and $f(\frac {2}{3}) = \frac {55}{27}$

Hence,

$f(1+\frac {2}{3}) = (1+\frac {2}{3}+1)(\frac {f(1)}{1+1} + \frac {f(\frac {2}{3})}{\frac {2}{3}+1} + (1)(\frac {2}{3}))$

$f(\frac {5}{3}) = (\frac {8}{3})(\frac {f(1)}{2} + \frac {f(\frac {2}{3})}{\frac {5}{3}} + \frac {2}{3})$

$f(\frac {5}{3}) = (\frac {8}{3})(\frac {4}{2} + \frac {\frac {55}{27}}{\frac {5}{3}} + \frac {2}{3})$

$f(\frac {5}{3}) = (\frac {8}{3})(2 + \frac {11}{9} + \frac {2}{3})$

$f(\frac {5}{3}) = (\frac {8}{3})(\frac {18}{9} + \frac {11}{9} + \frac {6}{9})$

$f(\frac {5}{3}) = (\frac {8}{3})(\frac {35}{9})$

$f(\frac {5}{3}) = \frac {280}{27}$

Because 280 and 27 are coprime positive integers, then a = 280 and b = 27 Thus, the value of a + b = 280 + 27 = 307

f(2/3)=(1/3+1/3)=(1+(1/3)/(1/3+1))f(1/3)+(1+(1/3)/(1/3+1))f(1/3)+1/27+1/9+1/27; simplifying it, we obtain: f(2/3)=(5/2)f(1/3)+5/27-----equation (1) since the question tells us that f(1)=4, we can rewrite it as f(1)=f(1/3+2/3)=4 from which we can get: f(1/3+2/3)=(1+(2/3)/(1/3+1))f(1/3)+(1+(1/3)/(2/3+1))f(2/3)+2/27+2/9+4/27; simplifying it, (3/2)f(1/3)+(6/5)f(2/3)+4/9=4------ equation (2) solve the two equations simultaneously, we get, f(1/3)=20/27, f(2/3)=55/27 since we are finding the value of f(5/3), which is equivalent to f(2/3+1) so f(5/3)=f(2/3+1) =(1+1/(2/3+1))f(2/3)+(1+(2/3)/(1+1))f(1)+4/9+2/3+2/3 =(8/5)f(2/3)+(4/3)f(1)+16/9 =(8/5)(55/27)+(4/3)(4)+16/9 = 280/27 hence a+b=280+27=307

P.S: I am not very familiar with the formatting methods since this is the first time I ever submit a solution. Sorry if the traditional way of presenting my answer causes you any trouble ;(

We wish to find $f(5/3)$ . Write this as $f(2-1/3)$ .

$\begin{array} {lcl} f(2+(-1/3)) & = & (1-1/9)f(2)+(1+3)f(-1/3)-4/3-2/3+2/9 \\ & = & 8f(2)/9+4f(-1/3)-16/9 \end{array}$

Calculate $f(2)$ as $f(1+1)=6+6+1+1+1=15$

$\begin{array} {lcl} f(2+(-1/3)) & = & 8f(2)/9+4f(-1/3)-16/9 \\ & = & 8*15/9+4f(-1/3)-16/9 \\ & = &104/9+4f(-1/3) \end{array}$

Now calculate $f(0)$ as $f(0+1)=f(1)$

$f(1)=2f(0)+f(1) \\ f(0)=0$

We now write $f(0)=f(1/3+(-1/3))=0$

$0 = 3f(1/3)/4+3f(-1/3)/2-1/9 \\ 2/3*(1/9-3f(1/3)/4)=f(-1/3) \\ 2/27-f(1/3)/2=f(-1/3)$

Now write $f(1/3+2/3)=f(1)=4$

$3f(1/3)/2+6f(2/3)/5+4/9=4\\ 3f(1/3)/2+6f(2/3)/5=32/9$

Now write $f(-1/3+2/3)=f(1/3)$

$2f(-1/3)+4f(2/3)/5-8/27=f(1/3)$

Now substitute the value we found for $f(-1/3)$

$4/27-f(1/3)+4f(2/3)/5-8/27=f(1/3) \\ -4/27+4f(2/3)=2f(1/3)$

Recall we found that:

$3f(1/3)/2+6f(2/3)/5=32/9$

So we have the system of equations:

$\begin{cases} -4/27+4f(2/3)=2f(1/3) \\ 3f(1/3)/2+6f(2/3)/5=32/9 \\ \end{cases}$

Going through the algebra and solving for $f(1/3)$ and $f(2/3)$ gives $f(1/3)=20/27$ and $f(2/3)=55/27$

We found:

$2/27-f(1/3)/2=f(-1/3)$

So substituting, we get

$2/27-10/27=-8/27=f(-1/3)$

Plugging this into $f(5/3)=104/9+4f(-1/3)$ , we get:

$f(5/3)=104/9-32/27=312/9-32/27=280/27$

So $a=280$ , $b=27$ , and $a+b=307$

Setting $x=y= \frac 13$ in the equation, we get, $f(\frac 23) = (1+ \dfrac {\frac 13} {1 + \frac 13}) f(\frac 13) +(1+ \dfrac {\frac 13} {1 + \frac 13}) f(\frac 13) + 2 (\frac 13)^3 + (\frac 13)^2$ $\Rightarrow f(\frac 23) = \frac 52 f(\frac 13) + \frac 5 {27}$

Again, setting $x = \frac 13, y = \frac 23$ in the equation, $f(1) = (1+ \dfrac {\frac 13} {1 + \frac 23}) f(\frac 23) +(1+ \dfrac {\frac 23} {1 + \frac 13}) f(\frac 13) +$ $\frac 13^2 \cdot \frac 23$ $+\frac 23 ^2 \cdot \frac 13 + \frac 23 \cdot \frac 13$

$\Rightarrow 4 = \frac 65 f(\frac 23) + \frac 32 f(\frac 13) + \frac49$

$\Rightarrow \frac{32} 9 = \frac 65 f(\frac 23) + \frac 32 f(\frac 13)$

Setting $f(\frac 23) = \frac 52 f(\frac 13) + \frac 5 {27}$ ,

$\frac{32} 9 = \frac 65 ( \frac 52 f(\frac 13) + \frac 5 {27} ) + \frac 32 f(\frac 13)$ $\Rightarrow \frac {32}{9} = \frac 92 f(\frac 13) + \frac 29 \Rightarrow f(\frac 13) = \frac 29 \cdot \frac {30}{9} = \frac {20}{27}$

So, $f(\frac 23) = \frac 52 \frac {20}{27} + \frac 5 {27} = \frac{55}{27}$

Finally, setting $x =1, y = \frac 23$ in the equation,

$f(\frac 53) = (1+ \dfrac {\frac 23} {1 +1}) f( 1) +(1+ \dfrac {1} {1 + \frac 23}) f(\frac 23) + 2 (\frac 23) + (\frac 23)^2$

$= \frac 43 \cdot 4 + \frac 85 \frac {55}{27} + \frac 83 +\frac 49$

$= \frac {16}3 + \frac {88}{27} + \frac {28}9 = \frac {280}{27}$

So, $\frac ab = \frac {280}{27}$ , and $a+b = 280+27 =307$

First we take $y=1$ . The given equation yields, $f(x+1)=\left(1+\frac{1}{x+1}\right)f(x) + 4+2x+x^2+x+x.$ That is, $f(x+1)=\left(\frac{x+2}{x+1}\right) f(x) + (x+2)^2.$ So, $\frac{f(x+1)}{x+2}=\frac{f(x)}{x+1}+x+2.$ Let $g(x)=\frac{f(x)}{x+1}.$ The above equation is equivalent to $g(x+1)=g(x)+x+2.$ Now, I claim that $g(x)=\frac{x^2+3x}{2}$ for $x \in \mathbb{Z}^+$ . We prove this by applying mathematical induction on $x.$ Clearly, the equation holds for $x=1.$ Now assume that it holds for some positive integer $x=n.$ Then, $g(n+1)=\frac{n^2+3n}{2}+n+2=\frac{(n+1)^2+3(n+1)}{2}.$ Thus, the equation for $g$ holds for all positive integers $x.$

In turn, this means that $f(x)=\frac{x(x+1)(x+3)}{2}.$ It follows that $f(\frac{5}{3})=\frac{280}{27}$ so our answer is $307.$

Note: The motivation for coming up with $g$ came from the resemblance of the recursive equation to that of triangle numbers.

Set $g(x) = \frac {f(x)}{x+1}$ . The equation simplifies to $g(x+y) = g(x) + g(y) + xy$ , where $g(1) = \frac{f(1)}{2} = 2$ . Set $h(x) = g(x) - \frac {x^2}{2}$ . The equation simplifies to $h(x+y) = h(x) + h(y)$ , where $h(1) = g(1) - \frac{1}{2} = \frac {3}{2}$ . By repeated addition, since $h(nx)=nh(x)$ this functional equation must satisfy $h \left(\frac {p}{q} \right) = p h \left(\frac {1}{q}\right) = \frac {p}{q} h(1)$ . Thus, for all rational numbers $r$ , we must have $g(r) = h(r) + \frac {r^2}{2} = \frac {r^2 + 3r}{2}$ , which gives $f(r) = (r+1) g(r) = \frac { r(r+1)(r+3) } {2}$ . Thus $f\left( \frac {5}{3}\right) = \frac {5}{3} \cdot \frac {8}{3} \cdot \frac {14}{3} \cdot \frac {1}{2} = \frac {280}{27}$ . Hence, $a+ b = 280 + 27 = 307$ .