Coefficients and Derivatives of Polynomials, and Solving $f' = f$ Hi so I have been spending a long time trying to figure these two problems out, 
1) Make up a few polynomials, express them as polynomials in $x-2$ and show that the coefficients are related to the derivatives at $2$.
2) Show that the only functions $f$ with $f' = f$ are $f(x) = e^x$
I think that for the first one I'm supposed to use some polynomials as an example to prove something, but I don't know how. And for the second one I believe I need to show constant multiples of $e^x$
 A: 1
\begin{alignat}{4}
f(x) &= b_0 + &b_1(x-2) + &b_2(x-2)^2 &+ \cdots + &\,b_n(x-2)^n &\qquad f(2) &= b_0\\
f'(x) &= &b_1 + &2b_2(x-2) &+ \cdots + &\,nb_n(x-2)^{n-1} &\qquad f'(2) &= b_1 \\
f''(x) &= &  &2b_2 &+ \cdots + &\,n(n-1)b_n(x-2)^{n-2} &\qquad f''(2) &= 2b_2 \\
\end{alignat}
The idea is, every time you differentiate, you get rid of the constant term, and every other term's degree is decremented by 1. When you plug in $2$, all the non-constant terms evaluate to $0$, so you are left with a constant term that is the result of the repeated differentiation of a single term. In general:
$$ f^{(m)}(2) = m! \cdot b_m $$
This is the essence behind Taylor polynomials. I recommend looking into it.
2
If we have a function $f(x)$ such that:
$$ f'(x) = f(x) $$
We see that:
\begin{align*}
\frac{d}{dx}\left[f(x)e^{-x} \right] &= \frac{d}{dx}\left[f(x)\right] \cdot e^{-x} + f(x) \cdot \frac{d}{dx}e^{-x} \\
&= f'(x)e^{-x} -f(x)e^{-x} \\
&= e^{-x}\left[f'(x)-f(x) \right] \\
&= e^{-x} \cdot 0 \\
&= 0
\end{align*}
Therefore, $f(x)e^{-x}$ is constant, meaning $f(x) = c e^{x}$ for some constant $c$.
A: The $2$nd part will be functions of the form $ce^x$ for any real constant $c$.
It's equivalent to solving the differential equation $dy/dx=y$.
