Let f(x)=xe^(-x). Prove that f^(n) (x)=(-1)^n * e^(-x) * (x-n) for every positive integer n.

I have no idea at all on how to do this.

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- Apr 18th 2007, 06:16 AMPossible actuaryno idea on this induction
Let f(x)=xe^(-x). Prove that f^(n) (x)=(-1)^n * e^(-x) * (x-n) for every positive integer n.

I have no idea at all on how to do this. - Apr 18th 2007, 06:22 AMPlato
Do you know how to differentiate that function?

Find the fourth derivative. Observe how the factors work.

Then you can do the induction. - Apr 18th 2007, 06:30 AMPossible actuary
I assume you mean to differentiate f(x)=xe^-x. Just curious, why the 4th derivative?

- Apr 18th 2007, 06:34 AMPlato
- Apr 18th 2007, 06:42 AMPossible actuary
I can see that the derivatives keep bouncing from xe^-x and -xe^-x + e^-x. So then I understand (-1^n) and (e^-x) but not (x-n).

- Apr 18th 2007, 06:59 AMecMathGeek
n = 0: f(x) = xe^(-x)

n = 1: f'(x) = x(-e^(-x)) + e^-x = -e^(-x)(x - 1)

n = 2: f''(x) = -e^(-x) + e^(-x)(x - 1) = e^(-x)(-1 + x - 1) = e^(-x)(x - 2)

n = 3: f'''(x) = e^(-x) - e^(-x)(x - 2) = e^(-x)(1 - x + 2) = -e^(-x)(x - 3)

n = 4: f''''(x) = -e^(-x) + e^(-x)(x - 3) = e^(-x)(-1 + x - 1) = e^(-x)(x - 4) - Apr 18th 2007, 07:07 AMecMathGeek
Here's how you show it inductively:

First, show that n = 1 works:

f'(x) = x(-e^(-x)) + e^-x = -e^(-x)(x - 1)

Next, assume it works for n = k:

f^{k}(x) = (-1)^(k)*e^(-x)(x - k)

Now, show it works for n = k + 1:

f^{k + 1}(x) = (-1)^(k + 1)*e^(-x)[x - (k + 1)]

Since f^{k + 1} = d/dx f^{k}, it holds that:

f^{k + 1}(x) = d/dx [(-1)^(k)*e^(-x)(x - k)]

= (-1)^(k)*e^(-x) - (-1)^(k)*e^(-x)(x - k)

= -1*(-1)^(k)*e^(-x)[-1 + x - k]

= (-1)^(k + 1)*e^(-x)[x - (k + 1)] = f^{k + 1}(x) - Apr 18th 2007, 10:59 AMThePerfectHacker
Try this with your induction skills.

Let f and g be infinitely differenciable at x.

Find a formula for the n-th derivative of (fg) , i.e. their product. - Apr 18th 2007, 03:31 PMecMathGeek
Are you asking me to try this?

(fg)' = fg' + f'g

(fg)'' = (fg' + f'g)' = fg'' + f'g' + f'g' + f''g = fg'' + 2f'g' + f''g

(fg)''' = (fg'' +2f'g' + f''g)' = fg''' + f'g'' + 2f'g'' + 2f''g' + f''g' + f'''g = fg''' + 3f'g'' + 3f''g' + f'''g

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.

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(fg)^{n} = C(n,0)*fg^{n} + C(n,1)*f'g^{n-1) + ... + C(n,n-1)*f^{n-1}*g' + C(n,n)*f^{n}g

Where C(n,i) is the combination (from probability theory) for the k'th term out of n terms, which is:

C(n,i) = n!/(i!*(n - i)!)

To prove this:

1. First, show it works for n = 1:

(fg)^{1} = C(1,0)*fg' + C(1,1)*f'g = fg' + f'g

2. Second, assume it's true for n = k

(fg)^{k} = C(k,0)*fg^{k} + C(k,1)*f'g^{k - 1} + ... + C(k,k-1)*f^{k-1}g' + C(k,k)*f^{k}g

3. Third, show it's true for n = k + 1

(fg)^{k + 1} = C(k+1,0)*fg^{k + 1} + C(k+1,1)*f'g^{k} + C(k+1,2)*f''g^{k - 1}... + C(k+1,k-1)*f^{k - 1}g'' + C(k+1,k)*f^{k}g' + C(k+1,k+1)*f^{k + 1}g

We have that (this is going to be a lot of writing):

(fg)^{k+1} = d/dx(fg)^{k}

= (C(k,0)*fg^{k+1} + C(k,0)*f'g^{k}) + (C(k,1)*fg^{k} + C(k,1)*f'g^{k-1}) + ... + (C(k,k-1)*f^{k-1}g'' + C(k,k-1)*f^{k}g') + (C(k,k)*f^{k}g' + C(k,k)*f^{k+1}g)

= [C(k,0)]*fg^{k+1} + [C(k,0) + C(k,1)]*f'g^{k} + ... + [C(k,k-1) + C(k,k)]*f^{k}g' + [C(k,k)]*f^{k+1}g

Now, it needs to be shown that C(k,i) + C(k,i+1) = C(k+1,i+1), where 0 <= i < k. I'll do this by direct proof.

Note that C(k,i) = k!/(i!*(k - i)!)

a. show it works for k = 1 (note that i = 0 since i + 1 = 1)

C(1,0) + C(1,1) = 1!/(0!*(1-1)!) + 1!/((0+1)!*(-0)!) = 1 + 1 = 2 = 2!/(1!(2-1)!) = C(2,1)

b. show it works for any value of k = j

C(j,i) + C(j,i+1) = j!/(i!(j - i)!) + j!/((i + 1)!(j - i - 1)!)

= j![(i + 1 + j - i)]/((i + 1)!(j - i)!)

= (j + 1)!/((i + 1)!((j + 1) - (i + 1))!)

= C(j+1,i+1) Q.E.D.

3. (continued) Therefore, the problem above becomes:

= [C(k,0)]*fg^{k+1} + [C(k,0) + C(k,1)]*f'g^{k} + ... + [C(k,k-1) + C(k,k)]*f^{k}g' + [C(k,k)]*f^{k+1}g

= C(k+1,0)*fg^{k+1} + C(k+1,1)*f'g^{k} + ... + C(k+1,k)f^{k}g' + C(k+1,k+1)f^{k+1}g

= (fg)^{k+1}

Q.E.D