Lower The 6th Degree

This is my 200th solution.

We shall use General Leibnitz Rule which states that

$\displaystyle \frac{d^n}{dx^n}\left(f g\right) = \sum_{k=0}^{n}{\binom{n}{k} f^{(k)} g^{(n-k)}}$

For $f = \ln(x), g = \frac{1}{x}$ , it becomes

$\displaystyle \frac{d^n}{dx^n}\left(\frac{\ln(x)}{x}\right) = \sum_{k=0}^{n}{\binom{n}{k} {\ln(x)}^{(k)} {\left(\frac{1}{x}\right)}^{(n-k)}}$

Now, we shall find ${\left(\frac{1}{x}\right)}^{(m)}$ for some integer $m$ .

Some simple computation shows that it equals $(-1)^k k! x^{-k-1}$ [You may use induction but that's trivial, however]

Now, what's ${\left(\ln(x)\right)}^{(m)}$ for some integer $m$ ? Well, $\ln(x)^{(1)} = \frac{1}{x}$ and then we know what's the nth derivative of $\frac{1}{x}$ . Hence, we can easily say that $\ln(x)^{(0)} = \ln(x), \ln(x)^{(m)} = (-1)^{m-1} (m-1)! x^{-m+1-1}, \forall m \in Z^+$

Substituting our general derivatives,

$\displaystyle \frac{d^n}{dx^n}\left(\frac{\ln(x)}{x}\right) =(-1)^n \binom{n}{0} n! \ln(x) x^{-n-1} + \sum_{k=1}^{n}{\binom{n}{k} (-1)^{k-1} (k-1)! x^{-k} (-1)^{n-k} (n-k)! x^{-n+k-1}}$

$\displaystyle = \frac{(-1)^n n! \ln(x)}{x^{n+1}} + \sum_{k=1}^{n}{\frac{n!}{k} \frac{(-1)^{n-1}}{x^{n+1}}}$

$\displaystyle = \frac{(-1)^n n! \ln(x) - (-1)^n n! H_n}{x^{n+1}}$ , $H_n$ is the Harmonic number.

The solution for $f^{(n)}(x) = 0$ hence becomes $e^{H_n}$ .

As a result, our final answer would be $e^{H_{10}}$ .

Where $H_n$ is the $n$ -th harmonic number,

$a_n=(-1)^nn!$

$b_n=H_n(-1)^nn!$

$\frac{a_nln(x)-b_n}{x^{n+1}} = \frac{(-1)^nn!}{x^{n+1}}(ln(x)-H_n)$

We can prove that $f^{(n)}(x) = \frac{(-1)^nn!}{x^{n+1}}(ln(x)-H_n)$ using induction.

Base Case: $f^{(0)}(x)=\frac{(-1)^0(0)!}{x^{0+1}}(ln(x)-H_0) = \frac{ln(x)}{x}$

$f^{(n+1)}(x) = \frac{d}{dx}\frac{(-1)^nn!}{x^{n+1}}(ln(x)-H_n)$

$f^{(n+1)}(x) = (-1)^nn!\left(\frac{x^n-(n+1)x^n(ln(x)-H_n)}{x^{2n+2}}\right)$

$f^{(n+1)}(x) = (-1)^{n+1}(n+1)!\left(\frac{(ln(x)-H_n)-\frac{1}{n+1}}{x^{n+2}}\right)$

$f^{(n+1)}(x) = \frac{(-1)^{n+1}(n+1)!}{x^{n+2}}\left(ln(x)-H_{n+1}\right)\ _\square$

Solving the equation $f^{(n)}(x)=0$ :

$\frac{(-1)^nn!}{x^{n+1}}(ln(x)-H_n)=0$

$ln(x)=H_n\Longrightarrow x=e^{H_n}$

Plugging in $n=10$ gives $x=e^{H_{10}}$ , so $p_{10}$ and $q_{10}$ must be the numerator and denominator of $H_{10}$ , respectively.

$H_{10} = \displaystyle\sum_{k=1}^{10} \frac{1}{k} = 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{10} = \frac{7381}{2520}$

Therefore $p_{10}=7381$ and $q_{10}=2520$ , so our answer must be:

$p_{10} + q_{10} = 7381 + 2520 = \boxed{9901}$