# Polynomials

We shall by computation discover that the polynomial functions

• $x_n(t) = \frac{t^n}{n!}$

for $n=0, 1, 2, 3,...$, where

• $t^n =t* t*...* t\quad$    (multiplication n times),
• $n!= n*(n-1)*(n-2)*...*2* 1$,

appear as solutions of the sequence of initial value problems for $n=1, 2, 3,...,$

• $Dx_n =x_{n-1}(t)$ (same as $dx_n =x_{n-1}dt$) for $t>0$,
• $x_n (0)= 0$.

where $D=\frac{d}{dt}$ denotes differentiation with respect to t and $x_0(t) = 1$.  We shall thus discover that

• $Dt^n = n*t^{n-1}$ for $n =1, 2, 3,...$

To check, we compute $x_n(t)$ successively for $n=1, 2, 3,...$ and verify by observing that blue and black curves agree as well the red line and black lines.

We have thus found by time stepping that the derivatives of the monomial polynomials $t^n = t*t*...*t$ (n times) are given by

• $\frac{dt^n}{dt}= n*t^{n-1}$ for $n=1, 2, 3,...$.

Summary: We have constructed the monomials $t^n = t*t*..*t$ as solutions to the differential equation  $Dt^n =n*t^{n-1}$. Or the other way around, we have shown that the derivative $Dt^n$ of $t^n=t*t*...*t$ is equal to $n*t^{n-1}$.

We have done this for $t>0$ and can extend to $t<0$ by time stepping backwards in time t. We can also let $t$ represent a spatial variable x and write

• $\frac{d}{dx}x^n = n*x^{n-1}$ for $n=1, 2, 3,...$.

We can finally extend to $n=-1,-2,-3,$ with $x\ne 0$, with a special rule for $n=0$ to be discovered as $\log(x)$.

To do:

1. Experiment with different time steps dt.

2. To see that $\frac{df}{dt} =2*t$ if $f(t)=t^2=t*t$, compute formally

• $df = f(t+dt)-f(t)=(t+dt)*(t+dt)- t*t= t*t+2*t*dt+dt*dt -t*t=2*t*dt + dt*dt$,

to find by dividing by dt

• $\frac{df}{dt} = 2*t + dt = 2*t$

if dt is vanishingly small (with a difference of size dt else). Of course also consider the case n=1 to (show that df/dt = 1 if f(t) = t).

3. Do the same computation for $f(t)=t^3=t*t*t$ and s0 on.

4. Compute the derivative of a polynomial $f(t) = a_0 + a_1*t + a_2*t^2 + a_3*t^3....+a_n*t^n$ with coefficients (numbers) $a_0, a_1,...,a_n$.