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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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.
Do you know how to differentiate that function?
Find the fourth derivative. Observe how the factors work.
Then you can do the induction.
I assume you mean to differentiate f(x)=xe^-x. Just curious, why the 4th derivative?
Well when you have done it four times, you should have some idea how the factors work. You should understand how the formula you trying to prove works.Quote:
Originally Posted by Possible 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).
n = 0: f(x) = xe^(-x)Quote:
Originally Posted by Possible actuary
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)
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)
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.
Are you asking me to try this?Quote:
Originally Posted by ThePerfectHacker
(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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(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