Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove the following identities:

$$ \left( x^n \ln x \right)^{(n)}= n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}$$

$$ \left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1} \left( \ln x - 1 - \frac{1}{2} - \cdots -\frac{1}{n} \right), \quad x>0,\quad n\ge 1, \tag{b}$$

Here $(\cdot )^{(n)}$ indicates the $n$-th derivative with respect to $x$.

I don't know where to start!

share|improve this question
2  
Is induction on $n$ not an option? –  Jyrki Lahtonen Dec 25 '11 at 11:41
    
For example you could use induction. Call $I_n (x)$ the RHside of the first formula; you can write the following recursion formula for $I_{n+1}(x)$: $$\tag{1} I_{n+1}(x)=(n+1)\ I_n (x) +n!\; .$$ Now, call $J(n)$ the LHside of the same formula: in order to prove $J_n (x)=I_n (x)$ for each $n\in \mathbb{N}$, it suffices to prove that $J_1 (x)=I_1 (x)$ and that also $J_n(x)$ satisfies recursion (1). I didn't make the computations, but I think this technique should work... If it doesn't, you could try with different recursion formulae. –  Pacciu Dec 25 '11 at 12:01
add comment

3 Answers

up vote 6 down vote accepted

I prefer using induction to prove the statement. For (a), when $n=1$, by product rule $$\left( x \ln x \right)'=\ln x+x\cdot\frac{1}{x}=1!(\ln x+1),$$ which shows that (a) is true for $n=1$. Now suppose that (a) is true for $n=k$, i.e. $$\left( x^k \ln x \right)^{(k)}= k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right).$$ Now, for $n=k+1$, we have $$\left( x^{k+1} \ln x \right)^{(k+1)}=\left[ \big(x^{k+1} \ln x\big)' \right]^{(k)}= \left[ (k+1)x^k \ln x+x^{k+1}\cdot\frac{1}{x}\right]^{(k)}$$ $$=(k+1)\left( x^k \ln x \right)^{(k)}+(x^k)^{(k)}.$$ Note that $(x^k)^{(k)}=k!$, by induction assumption, we have $$\left( x^{k+1} \ln x \right)^{(k+1)}=(k+1)k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right)+k!$$ $$=(k+1)! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k}+\frac{1}{k+1} \right),$$ which shows that (a) is true for $n=k+1$.

I will let you try (b) using induction.

share|improve this answer
add comment

To finish off Yuval's argument, we try an inductive proof on

$$\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}=-\sum_{k=1}^n \frac1{k}=-H_n$$

The equality is easily verified for $n=1$. We now try to simplify

$$\sum_{k=1}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}$$

which goes a bit like this:

$$\begin{align*} &\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \left(\binom{n+1}{k}-\binom{n}{k}\right)\frac{(-1)^k}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\sum_{k=1}^n (-1)^k\binom{n+1}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+\sum_{k=0}^{n+1} (1)^{n-k+1}(-1)^k\binom{n+1}{k}\right)\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+(1-1)^{n+1}\right)=-\frac1{n+1}\\ \end{align*}$$

and since we also have $-H_{n+1}-(-H_n)=-\dfrac1{n+1}$, the equality is established.


Here's an explicit application of Leibniz's product rule for the second problem, which uses the identity $\dfrac{\mathrm d^n}{\mathrm dx^n}\dfrac1{x}=\dfrac{(-1)^n n!}{x^{n+1}}$:

$$\begin{align*} \frac{\mathrm d^n}{\mathrm dx^n}\frac{\ln\,x}{x}&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\binom{n}{k}\left(\frac{\mathrm d^k}{\mathrm dx^k}\ln\,x\right)\left(\frac{\mathrm d^{n-k}}{\mathrm dx^{n-k}}\frac1{x}\right)\\ &=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\frac{n!}{\color{blue}{k!}\color{magenta}{(n-k)!}}\left(\frac{\color{red}{(-1)^{k-1}}\color{blue}{(k-1)!}}{\color{orange}{x^k}}\right)\left(\frac{\color{red}{(-1)^{n-k}}\color{magenta}{(n-k)!}}{\color{orange}{x^{n-k+1}}}\right)\\ &=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\frac{\color{red}{(-1)^{n-1}} n!}{\color{orange}{x^{n+1}}}\sum_{k=1}^n\frac1{\color{blue} k}=\frac{(-1)^n n!}{x^{n+1}}\left(\ln\,x-\sum_{k=1}^n\frac1{k}\right) \end{align*}$$

share|improve this answer
add comment

The familiar formula $(fg)' = f'g + fg'$ generalizes to the $n$th derivative in the following way: given a product of $m$ functions $f_1,\ldots,f_m$, the $n$th derivative is a sum of $m^n$ terms, each corresponding to a sequence of length $n$ listing which function is differentiated at every "step".

Write $x^n\ln x$ as a product of $n+1$ factors, $n$ of them equal to $x$ and the other one equal to $\ln x$. The $n$th derivative is a sum of terms, each corresponding to a sequence of length $n$. If any of the first $n$ functions appears twice, then the corresponding term is zero, since $x''=0$. So the general term lists $k$ different functions out of the first $n$, and the other one ($\ln x$) appears in the remaining $n-k$ positions. Call that a term of type $T_k$.

The term $T_n$ appears $n!$ times, and contributes $n!\ln x$ to the total sum. For $k < n$, the term $T_k$ appears $n(n-1)\cdots(n-k+1) \binom{n}{k}$ times. It is easy to prove by induction that for $m > 0$, $(\ln x)^{(m)} = (-1)^{m+1} (m-1)! x^{-m}$. Substituting $m = n-k$, we see that each appearance of $T_k$ contributes $(-1)^{n-k+1} (n-k-1)!$ to the sum. So all terms $T_k$ contribute $(-1)^{n-k+1} n!/(n-k) \binom{n}{k}$. Therefore the $n$th derivative is $$ \begin{align*} &n!\ln x + n!\sum_{k=0}^{n-1} (-1)^{n-k+1} \frac{1}{n-k} \binom{n}{k} \\ =&n!\ln x + n!\sum_{k=1}^n (-1)^{k+1} \frac{1}{k} \binom{n}{k}. \end{align*} $$ Now all that remains is to prove the identity $$ \sum_{k=1}^n (-1)^{k+1} \frac{1}{k} \binom{n}{k} = \sum_{k=1}^n \frac{1}{k}. $$

That should prove you with a good start on the first one. I trust that the second one is very similar (perhaps after substituting $y = 1/x$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.