Find $f''(x)$ if $f\circ f'(x) = 4x^2 + 3$

Can you tell me the solution of this question?

If: $f\circ f'(x)=4 x^2 +3$

then what is $f''(x)$?

This was a question in math test which I just took yesterday.

One function satisfying the equation above is $f(x)=x^2+3$, for which $f'(x)=2x$ and therefore $f''(x)=2$.

We can also see that $f'(x)$ is monotonic in $[0,+\infty)$, and in $(-\infty,0]$.

What other analytic solutions $f(x)$ exist ? Can we express all analytic solutions $f(x)$ with a few parameters ?

Solution 1:

we will show that the only smooth solution to the problem $f(f'(x)) = 4x^2 + 3 $ is $f(x) = x^2 + 3.$

the differentiable function $f$ is defined by $$ f(f^\prime(x)) = 4x^2 + 3.\tag 1$$ differentiating $(1)$ gives $$[f^\prime(f^\prime(x)) ] * f^{\prime \prime}(x)= 8x \tag 2 $$ differentiating $(2)$ gives $$[f^\prime(f^\prime(x))]*f^{\prime \prime \prime}(x) + \left( f^{\prime \prime} (x)\right)^2 f^{\prime \prime}(f^\prime(x))= 8 \tag 3$$ we will note that the range of $f$ is contained in $[3, \infty)$ and in fact $3$ is in the range of $f$ because putting $x = 0$ in $(1)$ gives $$f(f^\prime(0)) = 3, \ \ f^\prime(f^\prime(0)) = 0\tag 4$$ the second equality follows if we recognize that $3$ is a global minimum of $f.$

putting $x=0$ in $(3)$ and making use of $(4)$ shows that $f^{\prime \prime}(0) \neq 0.$ more is true. in fact we will show that $$ f^{\prime \prime}(x) \neq 0 \tag 5$$

suppose there is an $a$ such that $f^{\prime \prime}(a) = 0.$ putting $x = a$ in $(2)$ shows $a = 0.$ but we already showed that $f^{\prime \prime}(0) \neq 0$ and that establishes $(5).$ we can improve $(5)$ and will show that $$ f^{\prime \prime}(0) = 2,\ \ f^{\prime \prime}(x) > 0 \text { for all } x. \tag {5b} $$ this follows simply by putting $x = 0$ in $(3)$

we will now show that the only local extremum on the graph is $(0, 3)$

suppose $x=a$ is a local extremum. then by $(1)$ we have $a = \pm \sqrt{f(0) - 3}$. if these are distinct, then by rolle's thorem, there will be a point $c$ between them at which $f^{\prime \prime}(c) = 0$ contradicting $(5b).$ this tells us that $$f(0) = 3, f^\prime (0) = 0, f^{\prime \prime }(0) = 2, xf^\prime(x) > 0 \text{ for } x \neq 0, f^{\prime \prime}(x) > 0 \text{ for all } x. \tag 6$$

differentiating $(3)$ we get

$$[f^\prime(f^\prime (x)]*f^{(4)}(x) + f^{(3)}(x) f^{(2)}(f^\prime(x))f^{(2)}(x) + 2f^{(3)}(x) f^{(2)}(x)f^{(2)}(f^\prime(x) + f^{3}(f^\prime(x))(f^{(2)}(x))^3 = 0 \tag 7$$

observe that in both $(3)$ and $(7)$ the highest order derivative is multiplied by $[f^\prime(f^\prime(x))]$ which is zero at $x = 0$ it can be seen that this persists for all orders.

now putting $x = 0$ in $(7)$ and making use of $(6)$ gives us $4f^{(3)}(0) + 8 f^{(3)}(0) + 8f^{(3)}(0) = 0 $ which implies that $$f^{(3)}(0) = 0 $$

continuing this way i think i can show that $$f^{(n)}(0) = 0 \text{ for } n = 3, 4, 5, \cdots \tag 8$$ but it may be easier to expand $$f(x) = 3 + x^2 + 2a_4x^4 + 2a_5x^5 + \cdots $$

$$f^\prime(x) = 2x\left(1 + + 4a_4x^2 + 5a_5x^3 + \cdots \right) $$

putting these in $(1)$ we get

$$3 + 4x^2 = 3 + 4x^2 \left( 1 + 4a_4x^2 + 5a_5x^3 + \cdots \right)^2 + 2a_4(2x)^4 \left( 1 + 4a_4x^2 + 5a_5x^3 + \cdots \right)^4 + 5a_5 (2x)^5 \left( 1 + 4a_4x^2 + 5a_5x^3 + \cdots \right)^5+ \cdots $$

equating the coefficient of $x^4$ gives $a_4 = 0.$ now remove $a_4$ and show $a_5 = 0$ and by induction we should be able to show $(8)$.

p.s. thanks to the op for posting this problem.

the proof in the last stages can be streamlined using the following claim.

claim: if $f(x) = 3 + x^2 + 2ax^k, k \ge 3,$ then $a = 0.$

proof:

$\begin{align} 3 + 4x^2 &= f(f^\prime (x)) = f(2x(1 + akx^{k-2})) \\ &= 3 + 4x^2(1 +kax^{k-2})^2 + 2a*(2x)^k(1 + kax^{k-2})^k\\ &= 3 + 4x^2 + 8akx^k + 4k^2 a^2 x^{2k-2} + 2^{k+1}ax^k + \cdots \\ &= 3 + 4x^2 + a(8k + 2^{k+1})x^k + \cdots \end{align}$

which implies $ a = 0.$ this could be used to cut short the proof of $(8).$

Solution 2:

I think I have an (incomplete and not rigorous) heuristic which suggests that $f(x)=x^2+3$ is the only answer. Perhaps somebody can turn this into a proper/rigorous answer.

It starts with the assumption that $f$ is analytic:

$$f(x)=\sum_{i=0}^\infty a_ix^i,$$ so that $$f'(x)=\sum_{i=0}^\infty ia_ix^{i-1}.$$

If we write an expression for $f(f'(x))$ we get something like $$\begin{align} f(f'(x))&=a_0+a_1(a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots) \\&\phantom{=}+a_2(a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots)^2 \\&\phantom{=}+a_3(a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots)^3+\cdots \end{align}$$

If we can argue at this point (this is the big hole!) that all of the $a_n=0$ for $n\geq 3$ we are basically there. If this is true we have that $f(x)=a_0+a_1x+a_2x^2$. Just by plugging in we see that $a_2\neq 0$ and if $a_2\neq 0$ then $a_1=0$.

We plug $f(x)=a_0+a_2x^2$ into the relation and we are left with $a_0=3$ and $a_2^3=1\Rightarrow a_2=1$.