Convert Polynomial to Sum I have to convert the polynomial: $P_4=x^4+7x^3-13x^2-103x-84$ into the form: $$P_4(x)=\sum_{i=0}^4 a_i(x-1)^i$$ using Horner's Method.
I can evaluate both of them using Horner's Method, but I can't figure out how to convert them into one another.
 A: Alternative answer, using nothing but high-school algebra. 
I'm going to express $x^3 + 3x^1 - x + 4$ in powers of $x-1$. Why this polynomial? So that you can do the work with the real one. 
But why THIS one? I grabbed the coefficients randomly from thin air (keeping them fairly small to make the arithmetic not so bad). It turns out to work out reasonably well, but that was just blind luck, I promise. 
Let's go one degree at a time. 
$$
x^3 + 3x^2 -x + 4 = A(x-1)^3 + \text{lower order terms}.
$$
Now $(x-1)^3 = x^3 - 3x^2 + 3x + 1$, so $A$ times that is going to start with $Ax^3$...but we want $1 \cdot x^3$. So we have to pick $A = 1$. I'm going to write "LOT" for "lower order terms" from now on. Anyhow, we now know that $A$ has to be $1$, and we get
\begin{align}
x^3 + 3x^2 -x + 4 &= 1(x-1)^3 + LOT \\
x^3 + 3x^2 -x + 4 &= x^3 - 3x^2 + 3x - 1 + LOT \\
x^3 + 3x^2 -x + 4 - (x^3 - 3x^2 + 3x - 1) &=  LOT \\
 6x^2 -4x + 5  &=  LOT \\
\end{align}
and look! Now we have to write a quadratic in terms of powers of $(x-1)$. The first of those is going to be some multiple of $(x-1)^2$, so let's go!
\begin{align}
 6x^2 -4x + 5  &=  B(x-1)^2 + LOT \\
\end{align}
Now you can see that because $(x-1)^2 = x^2 -2x + 1$, that $B$ must be $6$. So write
\begin{align}
 6x^2 -4x + 5  &=  6(x-1)^2 + LOT \\
 6x^2 -4x + 5  &=  6(x^2-2x+1) + LOT \\
 6x^2 -4x + 5  &=  6x^2-12x+6 + LOT \\
  8x -1   &=  LOT \\
\end{align}
Now $8x-1$ is supposed to be a sum of powers of $(x-1)$, and it's clear the first of these is $C(x-1)$, and you can guess by now that $C = 8$. We get
\begin{align}
  8x -1   &=  8(x-1) + LOT \\
  8x -1   &=  8x-8 + LOT \\
   7   &=   LOT \\
\end{align}
and the final low-order term is $7(x-1)^0$. So we've got $A = 1, B = 6, C = 8, D = 7$, hence
$$
x^3 + 3x^2 -x + 4 = 1(x-1)^3 + 6(x-1)^2 + 8(x-1) + 7,
$$
assuming I did all the algebra correctly. Yay!
Now, can you repeat that process for your quartic? 
A: $$
P(x)=\sum_{i=0}^4 \frac{P^{(i)}(1)}{i!}(x-1)^i
$$
The best way to prove this identity (in general) is to use a little calculus. (Strictly speaking that's not necessary - there are finite difference arguments that don't call for the full machinery.) 
For your particular example, 
$$
P(1) = 1 + 7 -13 -103 - 84 = -192
$$
$$
P'(x) = 4x^3 + 21x^2 -26x - 103 
$$
$$
P'(1) = 4 + 21 -26 -103 = -104
$$
$$
P"(x) = P^{(2)}(x)= 12x^2 + 42x -26
$$
$$
P"(1) = P^{(2)}(1) = 12  + 42  -26 = 28
$$
$$
P^{(3)}(x) = 24x  + 42 
$$
$$
P^{(3)}(1) = 24  + 42 = 66
$$
$$
P^{(4)}(x) = 24  = P^{(4)}(1)
$$
so
$$
P(x) = -192 - 104(x-1) + \frac{28}{2}(x-1)^2 + \frac{66}{3 \cdot 2}(x-1)^3 + 
\frac{24}{4 \cdot 3 \cdot 2}(x-1)^4
$$
(if I haven't made any arithmetic errors).  The fact that the coefficients come out integers, as they must, is a preliminary check. You can verify for sure by multiplying it out.
A: Write $t=x-1$ and then calculate $$P_4(t+1)=(t+1)^4+7(t+1)^3-13(t+1)^2-103(t+1)-84$$ and you are done.
A: Note that:
$$a_0+a_1(x-1)+a_2(x-1)^2+a_3(x-1)^3+a_4(x-1)^4=x^4+7x^3-13x^2-103x-84 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_0}=1+7-13-103-84=\color{red}{-192};\\
(x-1)[a_1+a_2(x-1)+a_3(x-1)^2+a_4(x-1)^3]=x^4+7x^3-13x^2-103x+108 \stackrel{\div (x-1)}{\Rightarrow} \\
a_1+a_2(x-1)+a_3(x-1)^2+a_4(x-1)^3=x^3+8x^2-5x-108 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_1}=1+8-5-108=\color{red}{-104};\\
(x-1)[a_2+a_3(x-1)+a_4(x-1)^2]=x^3+8x^2-5x-4 \stackrel{\div (x-1)}{\Rightarrow} \\
a_2+a_3(x-1)+a_4(x-1)=x^2+9x+4 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_2}=1+9+4=\color{red}{14};\\
\vdots$$
Can you continue and finish?
