# Designing a polynomial

Can you find a polynomial function whose first derivative is equal to the function?

## Problem

**Can you design a polynomial in $x$, $p(x)$, such that $p(x)=p'(x)$?**

(a) What is the simplest polynomial in $x$ you can think of and what is its derivative?

(b) What powers of $x$ do you want the polynomial to include and what does this mean for the derivative?

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## Student Solutions

Thank you to Eddie from Wilson's School for submitting this very clear solution:

The simplest polynomial in $x$ is a linear one, such as $3x - 5$, and its

derivative is 3. Finding the derivative of a polynomial is simple: multiply

each term by its power in $x$, and reduce the power by 1. Since the power in

$x$ of every term is being reduced by 1, the derivative of the function will

never be equal to the function... for a finite-order function. If there is

no highest order term, i.e. the order is infinite, then it is possible.

For $p(x) = p'(x)$ to be true, each term $a_n x^n$ in $p(x)$ (where $a_n$ means "a

subscript n" and $a_n$ is the coefficient of $x^n$) must be equal to the

derivative of the next term $a_{n+1}x^{n+1}$ in p(x), which is

$(n+1) a_{n+1} x^n$. So:

$a_n x^n = (n+1) a_{n+1} x^n => a_n = (n+1) a_{n+1} =>

a_{n+1} = a_n/(n+1)$

So the coefficients of $x^n$ follow a sequence, such that the next

coefficient (that of $x^{n+1}$) is equal to the previous coefficient divided

by $n+1$. If we let $a_1 = 1$:

$a_1 = 1$, $a_2 = 1/2$, $a_3 = 1/6$, $a_4 = 1/24$ $=>$ $a_n = 1/n!$

So the polynomial is $p(x) = 1 + x + x^2/2 + x^3/6 + ... + x^n/n!+...$

This turns out to be the Maclaurin series expansion (a polynomial

approximation) of $e^x$, the simplest function such that $f(x) = f'(x)$.