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?