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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Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
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 $f(x) = f'(x)$.