1. ## Induction Proof

Hi,

Is this correct?

Prove by induction that $\displaystyle \frac{d^n}{dx^n}(xe^{-x}) = (-1)^n(x - n)e^{-x}$ for all positive integers n.

$\displaystyle \frac{d^n}{dx^n}(xe^{-x}) = (-1)^n(x - n)e^{-x}$

$\displaystyle \frac{d}{dx}(xe^{-x}) = -xe^{-x} + e^{-x}$

$\displaystyle = e^{-x(1 - x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$\displaystyle \,\,\,\,\,\,\,\,\,\,\text{...1} Prove true for n = 1 \displaystyle (-1)^n(x - n)e^{-x} \displaystyle (-1)^1(x - 1)e^{-x} \displaystyle (-1)(x - 1)e^{-x} \displaystyle (-x + 1)e^{-x}$$\displaystyle \,\,\,\,\,\,\,\,\,\,\text{...1}$

Prove true for n = k

$\displaystyle (-1)^n(x - n)e^{-x} = (-1)^k(x - k)e^{-x}$

Prove true for n = k + 1}

$\displaystyle (-1)^n\,(x - n)e^{-x}\,=\,(-1)^k\,+\,1[x - (k + 1)]e^{-x}$

$\displaystyle = (-1)^k\,+\,1(x - k - 1)e^{-x}$

$\displaystyle = (-1)^k\,(-1)^1\,(x - k - 1)e^{-x}$

$\displaystyle (-1)^{n}\,(x - n)e^{-x} = (-1)^{k}\,(-x + k + 1)e^{-x}$

2. Originally Posted by Hellbent
Hi,

Is this correct?

Prove by induction that $\displaystyle \frac{d^n}{dx^n}(xe^{-x}) = (-1)^n(x - n)e^{-x}$ for all positive integers n.

$\displaystyle \frac{d^n}{dx^n}(xe^{-x}) = (-1)^n(x - n)e^{-x}$

$\displaystyle \frac{d}{dx}(xe^{-x}) = -xe^{-x} + e^{-x}$

$\displaystyle = e^{-x(1 - x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$\displaystyle \,\,\,\,\,\,\,\,\,\,\text{...1} Prove true for n = 1 \displaystyle (-1)^n(x - n)e^{-x} \displaystyle (-1)^1(x - 1)e^{-x} \displaystyle (-1)(x - 1)e^{-x} \displaystyle (-x + 1)e^{-x}$$\displaystyle \,\,\,\,\,\,\,\,\,\,\text{...1}$

Prove true for n = k

$\displaystyle (-1)^n(x - n)e^{-x} = (-1)^k(x - k)e^{-x}$

Prove true for n = k + 1}

$\displaystyle (-1)^n\,(x - n)e^{-x}\,=\,(-1)^k\,+\,1[x - (k + 1)]e^{-x}$ this and the following lines are expressed incorrectly

$\displaystyle = (-1)^k\,+\,1(x - k - 1)e^{-x}$

$\displaystyle = (-1)^k\,(-1)^1\,(x - k - 1)e^{-x}$

$\displaystyle (-1)^{n}\,(x - n)e^{-x} = (-1)^{k}\,(-x + k + 1)e^{-x}$
$\displaystyle \displaystyle\frac{d^n}{dx^n}\left(xe^{-x}\right)=(-1)^n(x-n)e^{-x}$

Base Case

$\displaystyle n=1$

$\displaystyle \displaystyle\frac{d}{dx}\left(xe^{-x}\right)=e^{-x}-xe^{-x}=(1-x)e^{-x}=(-1)^1(x-1)e^{-x}$

P(k)

$\displaystyle \displaystyle\frac{d^k}{dx^k}\left(xe^{-x}\right)=(-1)^k(x-k)e^{-x}$

P(k+1)

$\displaystyle \displaystyle\frac{d^{k+1}}{dx^{k+1}}\left(xe^{-x}\right)=(-1)^{k+1}\left[x-(k+1)\right]e^{-x}$

Proof that P(k+1) will be valid if P(k) is

$\displaystyle \displaystyle\frac{d}{dx}(-1)^k(x-k)e^{-x}=(-1)^k\frac{d}{dx}(x-k)e^{-x}=(-1)^k\left[-(x-k)e^{-x}+e^{-x}(1)\right]$

$\displaystyle =(-1)^k(-1)\left[(x-k)e^{-x}-e^{-x}\right]=(-1)^{k+1}\left[x-(k+1)\right]e^{-x}$