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

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?

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 $$x$$