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Proof of the formula for the nth derivative of a product

  1. Feb 12, 2004 #1
    Someone please help me show that (uv)^n= Summation {u^(i)v^(n-i)}

    where i=0 the starting point and n is the ending number of the summation. By the way, there is also a term like (n) under the
    (i)
    summation which I wasn't sure how to account for. Since this is a more sophisticated version of induction which seems to account for two variables n and i, instead of just the usual n, I am a little stuck on how to approach this. Sorry for being a dumbass for the millionth time.
     
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  3. Feb 12, 2004 #2

    Hurkyl

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    Your goal is to prove it is true for all n, so it's your ordinary run of the mill induction style proof, not anything fancy. i is not a 'free' variable; it lives only inside the summation.

    P.S. I'm presuming you mean:

    [tex]
    \frac{d^n}{dx^n} (u v)
    = \sum_{i=0}^n \frac{d^i u}{dx^i} \frac{d^{n-i} v}{dx^{n-i}}
    [/tex]
     
  4. Feb 14, 2004 #3
    Yeah, you are correct in the way you wrote it Hurkyl. I'm still stuck however. I showed that the formula holds when n=0. And all I know about defining derivatives is given by the following. The oth derivative of w=w. The 1st derivative of w=w'. And the n+1 derivative of w = (the nth derivative of w)'. I know that the only trick to his problem has to do with changing the around the indices of the terms under summation, but I can't get to this point.
     
  5. Feb 14, 2004 #4

    Hurkyl

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    *shakes head* Bah, can't believe I missed that.

    I bet the formula you gave is wrong. :smile: Try computing by hand the case where n = 2.


    The right formula, and proof, looks almost exactly like the binomial theorem; have you seen a proof of that?
     
  6. Feb 14, 2004 #5
    Yeah, I have but that proof involved two free variables without involving summations. I think I have an idea of what to do. I know it can't be too too hard to solve. Thanks though.
     
  7. Feb 15, 2004 #6

    HallsofIvy

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    Try doing what Hurkyl suggested: look at a few cases first.


    for n= 1 (uv)'= u'v+ uv': the product rule

    for n= 2 (uv)"= ((uv)')'= (u'v+ uv')'
    = (u'v)'+ (uv')'= (u"v+ u'v')+ (u'v'+ uv")
    = u"v+ 2u'v'+ uv"

    Hmmm- that is NOT u"v+ u'v'+ uv" as your formula claims!

    for n= 3 (uv)"'= ((uv)")'= (u"v+ 2u'v'+ uv")'
    = (u"v)'+ 2(u'v')'+ (uv")'
    = (u"'v+ u"v')+ 2(u"v'+ u'v")+ (u'v"+ uv"')
    = u"'v+ 3u"v'+ 3u'v"+ uv"'

    Keep going until you recognize a pattern, then use induction to prove it.
     
  8. Feb 15, 2004 #7
    There is a term (n) under the summation, representing some integer.
    (i)
    Could the existence of this integer make it so that the formula I gave is correct?
     
  9. Feb 15, 2004 #8

    Hurkyl

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    The correct formula is

    [tex]

    \frac{d^n}{dx^n} (u v)

    = \sum_{i=0}^n \binom{n}{i} \frac{d^i u}{dx^i} \frac{d^{n-i} v}{dx^{n-i}}

    [/tex]

    Is this what you meant?
     
  10. Feb 15, 2004 #9
    Yes, that is what I mean, and how do you write in that lovely notation?
     
  11. Feb 17, 2004 #10

    HallsofIvy

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    If that was what you meant, you shouldn't have written something else! :smile: If you don't know how to use latex (that's what Hurkyl did) you might have used mCn for the binomial coefficient.

    In any case, click on the formula you are interested in and you will see the code used for it. Also check the "sticky posts" on Latex. It's tedious to use (imho) but looks very nice.
     
  12. Feb 29, 2004 #11
    In proving this nth derivative of a product formula, am I allowed to apply the theorem qn+1,r = qn,r-1 +
    qn,r (where all of these q terms are integers) to the summations?
     
  13. Feb 29, 2004 #12

    Hurkyl

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    If you know a theorem, then by all means use it. :smile: (Unless your teacher says otherwise)
     
  14. Mar 1, 2004 #13
    Ok, I been at it for a while, and I can't for the life of me prove this using induction.

    I am told the following
    u^0=u u^1=u' u^n+1= (u^n)'

    This and the laws for first derivative of a sum, and product, as well as dealing with multiplication by constants is all I know about differentiation.

    If P(n) is the statement regarding the nth derivative of a product
    formula, I cannot show that P(n) implies P(n+1).



    What I tried doing is differentiating the righthand side of the formula, and used the theorem that I stated two posts earlier to rewrite the left handside of the equation without success. I'm sorry if I am being confusing again, but can someone lead me in any direction at all?
     
  15. Mar 2, 2004 #14

    matt grime

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    proof by induction: clearly true for n=1 (we use [tex]\partial_x[/tex] to mean [tex]\frac{d}{dx}[/tex] to make it easier to write

    we assume [tex]\partial_x^n(uv)=\sum\binom{n}{r}\partial_x^r(u)\partial_x^{n-r}(v)[/tex]

    then

    [tex]\partial_x^{n+1}(uv)=\partial_x(\sum\binom{n}{r}\partial_x^r(u)\partial_x^{n-r}(v))[/tex]

    which is

    [tex]\sum\binom{n}{r]\partial_x(\partial_c^r(u)\partial_x^{n-r}(v))[/tex]

    here's the secret of proving things like this - notice I've not included the range of the sum - it takes care of itself always, so ignore it. do the differentiation


    [tex]\partial_x(\sum\binom{n}{r}(\partial_c^{r+1}(u)\partial_x^{n-r}(v)+\partial_x^{r}(u)\partial_x^{n-r+1}(v))[/tex]

    now work out the coefficient of [tex]\partial_x^s(u)\partial_x^{n+1-s}(v)[/tex] there are two contributions from different binomial coeffecients, and you can apply the result you wrote earlier on them.
     
    Last edited: Mar 2, 2004
  16. Mar 2, 2004 #15
    I'm sorry but I am not sure how you can just ignore the range of the sum. I don't see how it takes care of itself at all. What am I missing?
     
  17. Mar 2, 2004 #16

    matt grime

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    the binomial coefficients nCr are defined for all r, they are defined to be zero when r is not one of 0,1,..,n. and all the formulas you know are still true and it all works out. if you want you can carefully keep track of all the limits but you don't need to when summing over all binomial coefficients. it's one of the nicest 'cheats' there is in maths and if i'd spotted it (been told it) sooner it would have made life so much less messy.

    it's all down to how comfortable you are at switching labels in the indexing set - they are dummy variables after all.

    in the first term set s=r+1 and you get

    [tex]\binom{n}{r}\partial_x^r(u)\partial_x^{n-r}(v)=\binom{n}{s-1}\partial_x^s(u)\partial_x^{n+1-s}(v)[/tex]
    in the second one just put r=s and the whole thing is now

    [tex]\sum(\binom{n}{s-1}+\binom{n}{s})\partial_x^s(u)\partial_x^{n+1-s}(v) = \sum\binom{n+1}{s}\partial_x^s(u)\partial_x^{n+1-s}(v)[/tex]


    which is what you want
     
  18. Mar 2, 2004 #17
    Is that sufficient to show that P(n) implies P(n+1)? For some reason, I was working backwards and trying to cancel out terms so that we were left with the original statement P(n)
    [tex]\frac{d^n}{dx^n} (u v)= \sum_{i=0}^n \binom{n}{i} \frac{d^i u}{dx^i} \frac{d^{n-i} v}{dx^{n-i}}[/tex]

    You can just change the dummy variables whenever you want? How can we define r=s-1, and r=s in the same line of the equation? And can't we only use that binomial coefficient theorem when 1<=r<=n? Sorry for taking up so much of your time, and I really appreciate your help.
     
  19. Mar 2, 2004 #18

    matt grime

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    you can change the summation variable all you want, the binomial coeff relations are still true even when the r in n choose r is outside the range you're used to. you can check if you don't believe me. you can put the limits in and check in detail if you want but you don't get anything extra from it.

    what i wrote is the proof you want.

    i changed the variables like that without issue because i split the sum up into different parts and then glued it back together afterwards.

    imagine doing an integral of f(x)+g(x) over -inf to inf

    split it into int f plus int of g, and inside each do whatever subs you need to make it easier, it's like that.
     
  20. Mar 3, 2004 #19
    Thanks a million bro.
     
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