Establishing formulas using induction

Question

It is given that $$ \frac{d^n}{dx^n} (\frac{\ln x}{x})=\frac{a_n\ln x+b_n}{x^{n+1}}$$ where $a_n$ and $b_n$ depend only on $n$.

Use mathematical induction to establish a formula for $a_n$.

I tried differentiating the function but I obtained the recursive formula $a_{n+1}=-a_n(n+1)$.

Can somebody please provide some hints to solve this and also provide some mathematical details required to be stated in completing this proof?(I have never solved this type of a question so it would be helpful if a detailed answer is given).

@DonaldSplutterwit Thats okay, but how do I present this as a proof? — mathnoob123, May 20 '17 at 23:02
State the result $a_n=(-1)^{n} n!$ & claim that it is true. Then prove it by induction. — Donald Splutterwit, May 20 '17 at 23:03
Does anybody have an expression for $b_n$? $b_{n+1} = -b_n(n+1)+ a_n$ — caverac, May 20 '17 at 23:05
@caverac The question does not ask for a formula for $b_n$ ... the recurrence you state is correct, but an explicit formula will be difficult ? — Donald Splutterwit, May 20 '17 at 23:08
@DonaldSplutterwit I do understand the question doesn't ask for $b_n$, I am just curious. Thanks — caverac, May 20 '17 at 23:10
The first few values of $b_n=1,3,11,50,\cdots$ they occur in the second row of the Sterling numbers of the first kind. — Donald Splutterwit, May 20 '17 at 23:18

score 0 · Accepted Answer · answered May 27 '17 at 12:54

$$\frac{d^n}{dx^n} (\frac{\ln x}{x})=\frac{a_n\ln x+b_n}{x^{n+1}}$$ put n=1 $$\frac{d}{dx} (\frac{\ln x}{x})=\frac{\dfrac 1x .x-1.\ln x }{x^{2}}=\dfrac{-\ln x+1}{x^2}\to a_1=-1,b_1=+1$$ n=2 $$\frac{d^2}{dx^2} (\frac{\ln x}{x})=(\dfrac{-\ln x+1}{x^2})'=\\ \dfrac{-\frac 1x.x^2-2x(-\ln x+1)}{x^4}=\\ \dfrac{-1-2(\ln x+1)}{x^4}=\dfrac{+2\ln x-3}{x^3}\to a_2=+2$$

now suppose for $n$ that $\frac{d^n}{dx^n} (\frac{\ln x}{x})=\frac{a_n\ln x+b_n}{x^{n+1}}$ doing for $n+1$ $$\frac{d^{n+1}}{dx^{n+1}} (\frac{\ln x}{x})=\dfrac{d}{dx}\frac{d^n}{dx^n} (\frac{\ln x}{x})=\frac{a_n\ln x+b_n}{x^{n+1}}=\\\dfrac{d}{dx}\frac{a_n\ln x+b_n}{x^{n+1}}=\\ \dfrac{a_n\frac 1x.x^{n+1}-(n+1)x^n(a_n\ln x+b_b)}{x^{2n+2}}$$factor $x^n$ $$\dfrac{a_n-(n+1)(a_n\ln x+b_b)}{x^{n+2}}=\\ \dfrac{-(n+1)a_n\ln x+a_n-(n+1)b_n}{x^{n+2}}=\\ \dfrac{a_{n+1}\ln x+b_{n+1}}{x^{n+2}} \to a_{n+1}=-(n+1)a_n$$

Establishing formulas using induction

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