# Find value of this sum

Let $$\lim_{x\rightarrow 0}\frac{f^{}(x)}{x}=1$$

and for every $$x,y \in \mathbb{R}$$ we have:

$$f(x+y)=f(x)-f(y)+ xy(x+y)$$

Now Find : $$\sum_{i=11}^{17}f^{\prime} (i)$$

I think this question is false because we have:

Let $$x=y$$ then :

$$f(2x) =2x^3$$ and $$f(x)=\frac{x^3}{4}$$ But $$\lim_{x\rightarrow 0}\frac{f^{}(x)}{x}\neq1$$

• @amir bahadory: I think you are right, maybe a typo in question body exists Feb 29 at 19:54
• I think the intended approach was to rewrite the equation as $\dfrac{f(x+y) - f(x)}{y} = -\dfrac{f(y)}{y} + x(x+y)$ and take $y \to 0$ to obtain the derivative on the left and the RHS can be evaluated using the limit given. However, I do agree that there is an issue with the general functional equation. Feb 29 at 20:10
• Just based on the structure of the functional equation, I'm assuming it should have been something like $$f(x+y) = f(x) + f(y) + xy(x+y).$$ What is the source for this problem? If it's not your typo, it looks like a somewhat infamous error type caused by someone who knew the solution but not the theoretical requirements, so they fudged the details of the solution... Feb 29 at 20:11
• @BrianMoehring With the other functional equation $f(x+y) = f(x)+f(y)+xy(x+y)$ it works. Feb 29 at 21:07
• @psl2Z You'd have $g(x) = f(x) - \frac13x^3$ satisfies Cauchy's functional equation $g(x+y) = g(x) + g(y)$, which is well-known to have wildly non-continuous solutions if we allow the axiom of choice. Feb 29 at 23:37

There is no function $$f$$ with $$f(x+y) = f(x)-f(y)+xy(x+y)$$ for all $$x,y \in \mathbb{R}$$, since then with $$x = 0$$ it holds $$f(y) = f(0)-f(y)$$ for all $$y \in \mathbb{R}$$, so $$f =\frac{f(0)}{2}$$ constant. But then $$\frac{f(0)}{2} = xy(x+y)$$ for all $$x,y \in \mathbb{R}$$. A contradiction.

Suppose that $$f$$ instead satisfies $$f(x+y) = f(x)+f(y)+xy(x+y)$$ for all $$x,y \in \mathbb{R}$$. Then it holds with $$x= 0= y$$: $$f(0) = f(0) + f(0)$$, thus $$f(0) = 0$$, and with $$y = -x$$: $$0=f(0) = f(x)+f(-x)$$, thus $$f(-x) = -f(x)$$ for all $$x \in \mathbb{R}$$. Suppose $$f$$ is at least $$C^1$$. Then take the partial derivative $$\frac{\partial}{\partial x}$$ on both sides and get $$f'(x+y) = f'(x)+2xy+y^2$$ for all $$x,y\in \mathbb{R}$$. With $$x = 0$$ it follows $$f'(y) = f'(0)+y^2$$ for all $$y\in \mathbb{R}$$ and therefore $$f(y) = f'(0)y + \frac{1}{3}y^3 +c$$ for some $$c \in \mathbb{R}$$ and all $$y \in \mathbb{R}$$. It is $$f(0) = 0$$, so $$f(x) = f'(0)x + \frac{1}{3}x^3.$$

Now, define $$f(x):=ax+\frac{1}{3}x^3$$ for $$a, x \in \mathbb{R}$$. Then $$f(x+y) = a(x+y) + \frac{1}{3}(x+y)^3 = (ax + \frac{1}{3}x^3) +(ay + \frac{1}{3}y^3) + \frac{1}{3}(3x^2y+3y^2x) \\ = f(x)+f(y)+xy(x+y)$$ for all $$x,y \in \mathbb{R}$$, so satisfies the functional equation.

Thus all $$C^1$$ functions satisfying the functional equation for all $$x,y \in \mathbb{R}$$ are given by the family $$f_a(x) = ax+\frac{1}{3}x^3$$ for $$a, x \in \mathbb{R}$$.

We further know for an $$f$$ satisfying the functional equation that $$f$$ is $$C^1$$ if and only if the limit $$\lim_{h \to 0} \frac{f(h)}{h}$$ exists, i.e. $$f$$ is differentiable in $$0$$ (since $$f(0) = 0$$). This is because $$f(0) = 0$$ and from the functional equation it follows $$\frac{f(x+h)-f(x)}{h} = \frac{f(h)}{h}+x(x+h),$$ i.e. in this case $$f'(x) = a +x^2$$ for $$x \in \mathbb{R}$$ with $$a = \lim_{h \to 0}\frac{f(h)}{h}$$ (in particular, $$f'$$ is continuous).

Therefore, all functions $$f:\mathbb{R} \to \mathbb{R}$$ with existing limit $$\lim_{x \to 0} \frac{f(x)}{x}$$ satisfying the functional equation are given by $$f_a(x) = ax+\frac{1}{3}x^3$$ for $$x \in \mathbb{R}$$ with $$\lim_{x \to 0} \frac{f(x)}{x} = a \in \mathbb{R}$$.

Now assume $$\lim_{x \to 0} \frac{f_a(x)}{x} = 1$$. Then $$a = f_a'(0) = \lim_{x \to 0} \frac{f_a(x)}{x} = 1$$. Therefore $$f(x) = f_1(x) = x+\frac{1}{3}x^3.$$

It then follows: $$\sum_{i=11}^{17} f'(i) = \sum_{i=11}^{17} (1+i^2) = (17-10)+\frac{1}{6}(17(17+1)(2\cdot 17 +1) - 10(10+1)(2\cdot 10 +1)) = 7 +\frac{1}{6}(17\cdot 18 \cdot 35 - 10\cdot 11 \cdot 21) = 7 + 17 \cdot 3 \cdot 35 - 5 \cdot 11 \cdot 7 = 7+1785-385 = 7+1400 = 1407.$$