Given $a,b_0,\ldots,b_n$, there exists a polynomial of degree $\le n$ s.t. the derivatives $f^{(i)}(a)=b_i$ Just exploring some maths problems from a book until I came across this question. 
Let $a, b_0, . . . , b_n ∈ R$. Show that there exists a polynomial $f(x)$ of degree at most n such that
$$f(a) = b_0, f'
(a) = b_1, f''(a) = b_2,\ldots , f^{(n)}
(a) = b_n$$
I am not sure how to approach this problem, can someone give me a guide of how to do this problem? 

Consider the general polynomial of degree $n$
$$
f(x) = a_0 + a_1x + \ldots a_nx^n
$$
Since we know that a polynomial is uniquely determined by its Taylor expansion
, consider the taylor expansion of $f(x)$ about $a$, which is
$$
f(x) = f(a) + \frac{(x - a)f'(a)}{1!} + \frac{(x - a)^2f''(a)}{2!} + \ldots \frac{(x - a)^nf^n(x)}{n!}
$$
which gives us the unique polynomial in terms of the derivatives at $a$ and powers of $(x - a)$
