2
$\begingroup$

I have solved http://projecteuler.net/problem=44 by getting the inverse equation from Wikipedia http://en.wikipedia.org/wiki/Pentagonal_number:

Pentagonal: $f(n) = \frac{n(3n - 1)}{2}$

Inverse Pentagonal: $n = \frac{\sqrt{24f(n) + 1}+1}{6}$

am interested in the steps from Pentagonal equation (quadratic?) to the Inverse.

I note that it is similar to What is the inverse of $f(n)=\frac{n^2+n}{2}$? and I've tried the same strategy:

$f(n) = \frac{n(3n - 1)}{2}$

*6 + 1 on each side

$6f(n) + 1 = 9n^2 -3n + 1$

but this isn't correct because I want:

$6f(n) + 1 = 9n^2 -6n + 1$

to give:

$(3n-1)^2$ on the right hand side

  • 0
    You have multiplied by 6, try to multiply by 242012-06-29

2 Answers 2

1

To imitate the procedure used in the solution of the other problem, starting from $2f(n)=3n^2-n,$ multiply both sides by $12$. We get $24f(n)=36n^2-12n.$ Note that $36n^2-12n=(6n-1)^2-1$. The rest is easy. We get $(6n-1)^2=24f(n)+1$, then $6n-1=\sqrt{24f(n)+1}$, then $6n=\sqrt{24f(n)+1}+1$.

Another way to solve the same problem is to write our equation as $3n^2-n-2f(n)=0,$ and use the Quadratic Formula.

Remark: Look at the quadratic equation $ax^2+bx+c=0$, where $a\ne 0$. Multiply both sides by $4a$. We get the equivalent equation $4a^2x^2+4abx+4ac=0.$ Note that $4a^2x^2+4abx=(2ax+b)^2-b^2.$ So quickly we arrive at the equation $(2ax+b)^2=b^2-4ac.$ From this we conclude that $2ax+b=\pm\sqrt{b^2-4ac},$ and then straightforward algebra yields $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$ the important Quadratic Formula.

1

From $f(n)=n(3n-1)/2 \implies 3n^2-n=2f(n)\implies 3n^2-n-2f(n)=0$.This is a quadratic equation which gives $n=\frac{1+\sqrt{1+24f(n)}}{6}$.

  • 0
    Many thanks too avatar - the aha moment for me was getting the quadratic formula to give me 6 which then gave me the clue as to how you got 24.2012-06-29