Show that there is a unique differentiable function satisfying $f''(x)=a, f'(0)=b$, and $f(0)=c$.

Question

Suppose $a,b,c\in\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{R}$ is differentiable, $f''(x)=a \; \forall x$, $f'(0)=b$, and $f(0)=c$. Find $f$ and prove that it is the unique differentiable function with this property.

Obviously, $f(x)=\frac{a}{2}x^2+bx+c$. How can I show that this is the unique function with the above property? I was thinking of using Taylor's theorem and showing that the remainder is zero.

Since $f''(x)=a \; \forall x$, $f'''(x)=0$.

Apply Taylor's Theorem around $0$. There exists a $p$ between $x$ and $0$ such that $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(p)}{3!}x^3=c+bx+\frac{a}{2}x^2$.

Is this correct? Does this guarantee uniqueness? More generically, is there a one-to-one correspondence between a differentiable function and its Taylor expansion?

Answer 1

Yes, your proof is correct since your function is analytic and the Taylor series is equal to your function at any point.

Another approach is to assume two functions $f(x)$ and $g(x)$ satisfying the given conditions and proving that $g(x)=f(x)$ for every $x$

Note that $g''(x)=f''(x) =a \implies (g-f)''(x)=0$$

$$\implies (g-f)'(x)=C$$

Plug in $x=0$ and we get $C=0$, that is $g'(x)=f'(x)$ for all $x$

That and the condition on $g(0)=f(0)$ imply $g(x)=f(x)$ for all $x$