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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 $$\lim_{x\rightarrow 0}\frac{f^{}(x)}{x}\neq1$$

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    $\begingroup$ @amir bahadory: I think you are right, maybe a typo in question body exists $\endgroup$
    – Khosrotash
    Feb 29 at 19:54
  • $\begingroup$ 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. $\endgroup$
    – sudeep5221
    Feb 29 at 20:10
  • $\begingroup$ 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... $\endgroup$ Feb 29 at 20:11
  • $\begingroup$ @BrianMoehring With the other functional equation $f(x+y) = f(x)+f(y)+xy(x+y)$ it works. $\endgroup$
    – psl2Z
    Feb 29 at 21:07
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    $\begingroup$ @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. $\endgroup$ Feb 29 at 23:37

1 Answer 1

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

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