# Is this a correct way to prove what the derivative of a polynomial function is?

After trying a polynomial long division problem with a lot of wondering how to go about answering it I proceeded by most likely overcomplicating things but the equation derived seems to work at producing the derivative of a general polynomial function and I'm wondering whether it's a correct way of proving the result, even if it's rather long winded and inelegant!

If the polynomial is of the form: $$f(x)=(x-a)g(x)+r$$ where $r$ is just the remainder; now I went about trying to form an expression that was a general form for the process of polynomial long division. Starting with $$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$$ which I guessed would generate an expression for $g(x)$ in the form (this sum will also generate the remainder by letting the lower limit equal $0$): $$g(x)=\sum _{i=1}^{n-1} q_i\times x^{n-(i+1)}$$ where $q_i$ is the coefficient on the $n-(i+1)$ $x$ term. Now after doing some long division on the general terms of $f(x)$ it seemed that the coefficients could be calculated using the following: $$q(j,n,a)=\sum _{i=0}^{j-1} a_{(n-1)-i} a^{j-(i+1)}$$ where $j$ is linked to the index for the summation of $g(x)$ above (sorry if it's hard to follow, because I just guessed the summation that generates the coefficients the indices haven't worked out nicely), $n$ is the degree of $f(x)+1$, $a$ is the just the one from the factor $(x-a)$ and $a_i$ is the coefficient on the $x^i$ term in $f(x)$. We can combine the above two summations into one giving: $$g(x)=\sum _{i=1}^{n-1} q(i,n,a)\times x^{n-(i+1)}=\sum _{i=1}^{n-1} \left(x^{n-(i+1)} \sum _{j=0}^{i-1} a^{i-(j+1)} a_{(n-1)-j}\right)$$

you can see why I said it was long winded and inelegant now! But if everything is correct the expression above generates $g(x)$ given $f(x)$ and what it's being divided by: $(x-a)$.

Now if we look at the expression $f(x)=(x-a)g(x)+r$ again we can use the remainder theorem to rewrite the remainder as: $r=f(a)$. Therefore giving: $$f(x)=(x-a)g(x)+f(a)\longrightarrow \frac{f(x)-f(a)}{(x-a)}=g(x)$$ From here it's in the form that kind of resembles the definition of a derivative. Therefore if we took the limit as $x$ approached $a$ : $$\mathop{\lim}\limits_{x \to a} \frac{f(x)-f(a)}{(x-a)}=f'(a)=g(a)$$ Thus if we applied this to the equation derived above we would get: $$f'(a)=\mathop{\lim}\limits_{x \to a} \sum _{i=1}^{n-1} \left(x^{n-(i+1)} \sum _{j=0}^{i-1} a^{i-(j+1)} a_{(n-1)-j}\right)=\sum _{i=1}^{n-1} \left(a^{n-(i+1)} \sum _{j=0}^{i-1} a^{i-(j+1)} a_{(n-1)-j}\right)$$ and then after a little simplification we can get the equation:

$$f'(a)=\frac{1}{a^2}\sum _{i=1}^{n-1} a^{n-i} \sum _{j=0}^{i-1} a^{i-j} a_{(n-1)-j}$$ Now rewriting the above expression in terms of $x$ so it gives the derivative of the polynomial $f(x)$ with coefficients $a_i$ at any point $x$ as: $$f'(x)=\frac{1}{x^2}\sum _{i=1}^{n-1} x^{n-i} \sum _{j=0}^{i-1} x^{i-j} a_{(n-1)-j}$$ To see if it worked I wrote a piece of Mathematica code that would carry the above formula out and it indeed does seem to work, the code is below with $n$ as the degree of $f(x)+1$ so for a cubic $n=4$.

test[n_] := Collect[1/x^2* Sum[x^(n - (i)) Sum[ Subscript[a, (n - 1) - j]*x^(i - (j)), {j,0, i - 1}], {i, 1, n - 1}], x]

Now after writing all this and looking up at the mathematical mess I'm still curious to know whether it's a valid way of proving what the derivative of a general polynomial is since it seems to work but I would have expected something more in the form $ix^{i-1}$ but I can't see any obvious way of simplifying it down into that form. If you do use the Mathematica code you can delete the Collect function and see that the $i$ appears because the $x^{i-1}$ term is in the expression $i$ times but I can't see how to get it into just one summation. I apologise if my explanation isn't the clearest and hopefully I didn't make any stupid mistakes!

• Interesting process. It appears to be sound. – John Habert Feb 3 '14 at 4:55
• You might want to check the algorithm and proof of correctness of the Horner scheme and Ruffini rule. Which is essentially the computation you performed. – LutzL Feb 3 '14 at 15:03
• The way I think about it is that the derivative quotient removes all terms in a polynomial after binomial expansion of $x+h$ because all things left with h end up becoming $0$ and the $x^n$ term is canceled. – Simply Beautiful Art Dec 16 '15 at 0:38