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? 
 A: 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 $(x - a)$
A: Consider the following polynomial:
$$ p(x) = \sum\limits_{i=0}^n \frac{b_i}{i!}(x-a)^i $$
And observe that:
$p^{(k)}(a) = \left(\sum\limits_{i=0}^{k-1} 0\right) + \sum\limits_{i=k}^n\frac{b_i}{(i-k)!}(x-a)^{i-k} = b_k $
A: I'll show it for a cubic polynomial, let $f(x)=px^3+qx^2+rx+s$.
Expressing the given constraints, we have
$$\begin{cases}f(a)&=pa^3+qa^2+ra+s\\f'(a)&=3pa^2+2qa+r\\f''(a)&=6pa+2q\\f'''(a)&=6p.\end{cases}$$
This is a system of linear equations in triangular form, which can be readily solved.
$$\begin{cases}p=\frac16f'''(a)\\q=\frac12(f''(a)-6pa)\\r=f'(a)-3pa^2-2qa\\s=f(a)-pa^3-qa^2-ra.\end{cases}$$
As the diagonal coefficients are never zero, a solution is always guaranteed. You can generalize for any degree.
