Find a polynomial with conditions on its derivatives Find a poly of degree $4$ such that
$$f(2)=5
\\f'(2)=19
\\f''(2)=40
\\f^{(3)}(2)=48
\\f^{(4)}(2)=24$$
Where to start?
My approach:
I pick a basis $\{f,f',f'',f^{(3)},f^{(4)}\}$ then the action of evaluation of poly at 2  on a $\{e_i|i=1,..,4\}$ vectors would give as ${5, 19,40, 48, 24}$  respectively and what?
Edit:
Thanks to @Fred and @dxiv for a Taylor series hint. However I would like to see how one can use linear algebra to solve it, so if anyone has some idea please post it.
 A: Recall that $\{1, x-2, (x-2)^2, (x-2)^3, (x-2)^4\}$ is a basis for the space $\mathbb{R}_{\le 4}[x]$.
Therefore $$f(x) = a_0 + a_1(x-2) + a_2(x-2)^2 + a_3(x-2)^3 + a_4(x-2)^4$$
We have $a_0 = f(2) = 5$.
Taking the derivative gives
$$19 = f'(2) = \Big(a_1 + 2a_2(x-2) + 3a_3(x-2)^2 + 4a_4(x-2)^3\Big)\Bigg|_{x=2} = a_1$$
Again 
$$40 = f''(2) = \Big(2a_2 + 6a_3(x-2) + 12a_4(x-2)^2\Big)\Bigg|_{x=2} = 2a_2$$
so $a_2 = 20$.
Continue inductively in this manner, you will get $f^{(k)}(2) = k! a_k$.
A: We have $f(x)= \sum_{k=0}^4 \frac{f^{(k)}(2)}{k!}(x-2)^k.$
Can you proceed ?
A: The coefficients of any polynomial are the valuations of the derivatives at $0$, divided by the factorial of the order of the derivative*.
$$p_k=\frac{P^{(k)}(0)}{k!}.$$
So if you consider the polynomial 
$$P(x):= 5+19x+\frac{40}2x^2+\frac{48}6x^3+\frac{24}{24}x^4,$$ it fits the given data around $x=0$.
Now by translation,
$$Q(x):=P(x-2)$$ fits around $x=2$.

*To illustrate the above property,
$$(a+bx+cx^2+dx^3+ex^4)'''=3\cdot2\cdot1\cdot d+4\cdot3\cdot2\cdot ex$$ gives $3! d$ when evaluated at $0$.
A: Taylor series:
$$f(x)=f(t)+f'(t)(x-t)+\frac{f''(t)}{2!}(x-t)^2+\frac{f'''(t)}{3!}(x-t)^3+\frac{f''''(t)}{4!}(x-t)^4$$
where your $t$ is $2$
We know our polynomial has the general formula:
$$f(x)=ax^4+bx^3+cx^2+dx+e$$
so we can obtain the following equations:
$$16a+8b+4c+2d+e=5$$
$$32a+12b+4c+d=19$$
$$48a+12b+2c=40$$
$$48a+6b=48$$
$$24a=24$$
each of these has come from the correct derivative of the equation, then substitution $x=2$
we know that $a=1$, so $6b=0$ therefore $b=0$
we now have $48+2c=40$ so we know $c=-4$
we now have $16+d=19$ so we know $d=3$
we now have $6+e=5$ so we know $e=-1$
so to recap: $$a=1,b=0,c=-4,d=3,e=-1$$
so our polynomial has the expression:
$$f(x)=x^4-4x^2+3x-1$$
