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Given that $ U_n=[x(1-x)]^n$ and $n\geq2$,$V_n=\int_{0}^{1}e^xU_ndx$,prove that $V_n+2n(2n-1)V_{n-1}-n(n-1)V_{n-2}=0$

I tried to solve it by integration by parts,taking $U_n$ as first function and $e^x$ as second function.

$V_n=-\int_{0}^{1}n[x(1-x)]^{n-1}(1-2x)e^x dx=-\int_{0}^{1}nU_{n-1}(1-x-x)e^x dx$

I could not proceed further,how can i reach the desired result.Please help.

Brahmagupta
  • 4,204

2 Answers2

1

You can integrate $U_n \cdot e^x$, $U_{n-1} \cdot e^x$ and $xU_{n-1} \cdot e^x$ by parts then look for a suitable linear combination. Alternatively, you can find directly that $$ (U_n +2n(2n-1)U_{n-1}-n(n-1)U_{n-2})\cdot e^x = \Big((U_n-n(1-2x)U_{n-1})\cdot e^x\Big)'. $$

G.Kós
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1

The factors $n(n-1)$ and $2n(2n-1)$ indicate that we could try integration by parts twice.

We obtain for $n\geq 2$ \begin{align*} V_n&=\int_0^1e^xU_n(x)dx\\ &=\int_0^1e^xx^n(1-x)^ndx\\ &=\left.e^xx^n(1-x)^n\right|_0^1-\int_0^1e^x\left[nx^{n-1}(1-x)^n-nx^n(1-x)^{n-1}\right]dx\tag{1}\\ &=-n\int_0^1e^xx^{n-1}(1-x)^ndx+n\int_0^1e^xx^{n}(1-x)^{n-1}dx\\ &=n\int_0^1e^x\left[(n-1)x^{n-2}(1-x)^n-nx^{n-1}(1-x)^{n-1}\right]dx\tag{2}\\ &\qquad-n\int_0^1e^x\left[nx^{n-1}(1-x)^{n-1}-(n-1)x^n(1-x)^{n-2}\right]dx \end{align*}

Comment:

  • In (1) and (2) we do integration by parts

The next step is to collect the summands in (2). Since the integral is a linear operator we can focus on the polynomial parts. We also conveniently use $U_n(x)$ as shorthand for $x^n(1-x)^n$.

We obtain from (2)

\begin{align*} n&(n-1)(1-x)^2U_{n-2}(x)-n^2U_{n-1}(x)-n^2U_{n-1}(x)+n(n-1)x^2U_{n-2}(x)\\ &=n(n-1)(1-2x+2x^2)U_{n-2}(x)-2n^2U_{n-1}(x)\\ &=n(n-1)(1-2x(1-x))U_{n-2}(x)-2n^2U_{n-1}(x)\\ &=n(n-1)U_{n-2}(x)-2n(n-1)U_{n-1}(x)-2n^2U_{n-1}(x)\\ &=n(n-1)U_{n-2}(x)-2n(2n-1)U_{n-1}(x)\tag{3} \end{align*}

Now it's time to harvest.

We obtain with (2) and (3) \begin{align*} V_n&=-2n(2n-1)\int_0^1e^xU_{n-1}(x)\,dx+n(n-1)\int_{0}^1e^xU_{n-2}(x)\,dx\\ &=-2n(2n-1)V_{n-1}+n(n-1)V_{n-2} \end{align*}

and the claim follows.

Markus Scheuer
  • 108,315