mxchickmagnet86
New member
- Joined
- Dec 6, 2008
- Messages
- 3
*post has been deleted by user*
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Proof: [(d^n)(x e^(-x))]/(dx)^n = (-1)^n (x - n)e^(-x)
- Thread starter mxchickmagnet86
- Start date
Re: Proofs Help!
\(\displaystyle \frac{d^{n}}{dx^{n}}[xe^{-x}]=(-1)^{n}(x-n)e^{-x}\)
Note that in Leibniz's rule, the indices of the derivatives are the same as the exponents of the binomial theorem.
Use \(\displaystyle u=x, \;\ v=e^{-x}\)
\(\displaystyle \frac{du}{dx}=1, \;\ \frac{dv}{dx}=-e^{-x}\)
\(\displaystyle \frac{d^{2}u}{dx^{2}}=0, \;\ \frac{d^{2}v}{dx^{2}}=e^{-x}\)
Note, after we get n>1 we have \(\displaystyle \frac{d^{n}u}{dx^{n}}=0\)
That creates cancellations in the general expansion that simplify things.
Here's a general form of Leibniz's rule for products. You can probably find it in a book or on the web somewhere.
Use it for your case. Sub in and simplify.
\(\displaystyle \frac{d^{n}}{dx^{n}}(uv)=\frac{d^{n}u}{dx^{n}}v+n\frac{d^{n-1}u}{dx^{n-1}}\frac{dv}{dx}+\frac{n(n-1)}{2!}\frac{d^{n-2}u}{dx^{n-2}}\frac{d^{2}u}{dx^{2}}+.......\)
\(\displaystyle =\sum_{k=0}^{n}C(n,k)\right)\frac{d^{k}}{dx^{k}}u\cdot \frac{d^{n-k}}{dx^{n-k}}v\)
Because of the nth derivatives of u, we are left with \(\displaystyle (-1)^{n}\left[xe^{-x}-ne^{-x}\right]\)
\(\displaystyle \frac{d^{n}}{dx^{n}}[xe^{-x}]=(-1)^{n}(x-n)e^{-x}\)
Note that in Leibniz's rule, the indices of the derivatives are the same as the exponents of the binomial theorem.
Use \(\displaystyle u=x, \;\ v=e^{-x}\)
\(\displaystyle \frac{du}{dx}=1, \;\ \frac{dv}{dx}=-e^{-x}\)
\(\displaystyle \frac{d^{2}u}{dx^{2}}=0, \;\ \frac{d^{2}v}{dx^{2}}=e^{-x}\)
Note, after we get n>1 we have \(\displaystyle \frac{d^{n}u}{dx^{n}}=0\)
That creates cancellations in the general expansion that simplify things.
Here's a general form of Leibniz's rule for products. You can probably find it in a book or on the web somewhere.
Use it for your case. Sub in and simplify.
\(\displaystyle \frac{d^{n}}{dx^{n}}(uv)=\frac{d^{n}u}{dx^{n}}v+n\frac{d^{n-1}u}{dx^{n-1}}\frac{dv}{dx}+\frac{n(n-1)}{2!}\frac{d^{n-2}u}{dx^{n-2}}\frac{d^{2}u}{dx^{2}}+.......\)
\(\displaystyle =\sum_{k=0}^{n}C(n,k)\right)\frac{d^{k}}{dx^{k}}u\cdot \frac{d^{n-k}}{dx^{n-k}}v\)
Because of the nth derivatives of u, we are left with \(\displaystyle (-1)^{n}\left[xe^{-x}-ne^{-x}\right]\)
Re: Proofs Help!
One can do this by induction as well. for n=1 we have d/dx [ xe^(-x) ] = -xe^(-x) + e^(-x) = e^(-x)*(-x+1) = -e^(-x)(x-1)
assume for all k less than or equal to a fixed n that P(k) holds. i.e. P(n) holds.
then d^n/dx^n (xe^(-x)) = (-1)^n*(x-n)*e^(-x)
now take the derivative (the [n+1]th derivative of f) using the product rule, and do a bit of factoring, to complete the proof. note (-1)^n is a constant number (we fixed n).
One can do this by induction as well. for n=1 we have d/dx [ xe^(-x) ] = -xe^(-x) + e^(-x) = e^(-x)*(-x+1) = -e^(-x)(x-1)
assume for all k less than or equal to a fixed n that P(k) holds. i.e. P(n) holds.
then d^n/dx^n (xe^(-x)) = (-1)^n*(x-n)*e^(-x)
now take the derivative (the [n+1]th derivative of f) using the product rule, and do a bit of factoring, to complete the proof. note (-1)^n is a constant number (we fixed n).