1

A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If f(1)=2, what is f(8)?

I would like to try to tackle this problem but I need somewhere to start as I really have no idea at all on how to start.

At first I assumed that perhaps $f$ was a polynomial function, like a quadratic, but the $3xy$ term in the equation says otherwise.

Update -Thanks for everyone's help

This is what everyone has suggested. Instead of trying to find out the function, just use the f(1) value to find f(8):

$f(1+1)= f(2) = f(1) + f(1) +3(1)(1) = 7$ $f(4) = f(2) + f(2) + 3(2)(2) = 26$ $f(8) = f(4) + f(4) + 3(4)(4) = 100$

user2574069
  • 51
  • 4
    There is no property mentioned. You want an equality sign somewhere. – André Nicolas Apr 18 '14 at 23:12
  • 1
    Did you mean $f(x+y)\boxed{=}f(x)+f(y)+3xy$? –  Apr 18 '14 at 23:16
  • So, suppose $f(x+y) = f(x) + f(y) + 3xy$. Then, $f(2) = 2f(1) + 3 = 7$. Then, $f(4) = 2f(2) + 12 = 26$ and finally $f(8) = 2f(4) + 48 = 100$. Oops, you said you wanted to do it. I kind of ruined it? – Christopher K Apr 18 '14 at 23:23
  • Yes sorry typo with the equality sign – user2574069 Apr 18 '14 at 23:24
  • 1
    Your edit gives an incorrect answer - you're adding $3f(x)f(y)$, not $3xy$. – Steven Stadnicki Apr 18 '14 at 23:28
  • I suggest that we actually determine the function so that we know it is well defined. For example, we know that $f(2)=f(1)+f(1)+3\cdot1\cdot1=7$ and that $f(3)=f(1)+f(2)+3\cdot1\cdot2=15$. Now we can compute $f(4)$ in two ways: $f(3)+f(1)+3\cdot3\cdot1=26$ and $f(2)+f(2)+3\cdot2\cdot2=26$. To make sure this happens in every case, we should determine the function. – robjohn Apr 18 '14 at 23:48

4 Answers4

1

Assuming you mean $f(x+y)=f(x)+f(y)+3xy$.

Hint. Taking $x=y=1$, we have $$f(2)=f(1)+f(1)+3=7$$ and so on.

David
  • 82,662
1

Let me list out a table, assuming you mean $f(x+y)=f(x)+f(y)+3xy$: \begin{array}{|c|c|} x & f(x)\\ 1& 2\\ 2 & 7\\ 4 & 26\\ 8 & \boxed{100} \end{array}

1

Setting $y=x$, we get $$f(2x) = 2f(x) + 3x^2$$ Hence, if $x=2^n$, we get $$f(2^{n+1}) = 2f(2^n) + 3 \cdot 2^{2n} = 2(2f(2^{n-1}) + 3 \cdot 2^{2n-2}) + 3 \cdot 2^{2n} = 4 f(2^{n-1}) + 3(2^{2n-1} + 2^{2n})$$ In general, we have $$f(2^{n+1}) = 2^{k+1} f(2^{n-k}) + 3 \left(2^{2n} + 2^{2n-1} + \cdots +2^{2n-k}\right) = 2^{k+1}f(2^{n-k}) + 3 \cdot 2^{2n-k} \left(2^{k+1}-1\right)$$ Setting $k=n$, we get $$f(2^{n+1}) = 2^{n+1} f(1) + 3 \cdot 2^n \cdot \left(2^{n+1}-1\right)$$ Setting $n=2$, gives you the answer.

A more direct way is to recognize that $$f(x+1) = f(x) + f(1) + 3x$$ Similarly, we get \begin{align} f(x) & = f(x-1) + f(1) + 3(x-1)\\ f(x-1) & = f(x-2) + f(1) + 3(x-2)\\ f(x-2) & = f(x-3) + f(1) + 3(x-3)\\ \vdots & = \vdots\\ f(2) & = f(1) + f(1) + 3 \cdot 1 \end{align} Using telescopic summation we get $$f(x+1) = f(1) + x f(1) + 3 \sum_{k=1}^x k$$ Hence, $$f(x+1) = (x+1)f(1) + \dfrac{3x(x+1)}2$$ Since $f(1)$ is given as $2$, we get $$f(x+1) = 2(x+1) + \dfrac{3x(x+1)}2 = \dfrac{(x+1)(3x+4)}2$$ Hence, $$f(x) = \dfrac{x(3x+1)}2$$

user141421
  • 2,538
1

We can also set $y=1$ and solve the recursion $$ f(x)=f(x-1)+3x-1 $$ to get that for $x\in\mathbb{Z}$, $$ \begin{align} f(x) &=\sum_{k=1}^x3k-1\\ &=x(3x+1)/2 \end{align} $$ Checking this formula against the given relation, we see that $f$ is well-defined (it doesn't depend on how we get to a particular argument): $$ (x+y)(3(x+y)+1)/2=x(3x+1)/2+y(3y+1)/2+3xy $$

Calculate $f(8)$: $$ f(8)=8(3\cdot8+1)/2=100 $$

robjohn
  • 345,667