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 October 31st, 2017, 10:40 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 205 Thanks: 26 n-th derivative Check if this holds true .... (post your check) N-th derivative example without inspection of first , second , third ... derivative $\displaystyle y=xe^x \;$ $$(uv)^{(n)}= \sum \limits_{k=0}^{n} {n \choose k} u^{(n-k}v^{(k)} \;$$ set $\displaystyle u=x , v=e^x$ $$y^{(2n)}= \sum \limits_{k=0}^{2n} {2n \choose k} x^{(2n-k)}(e^x)^{(k)}=e^x \sum \limits_{k=0}^{2n} {2n \choose k} x^{(2n-k)}= {2n \choose 0} x^{(2n)}+{2n \choose 1} x^{(2n-1)}+...+{2n \choose 2n-1} x^{(1)}+{2n \choose 2n} x^{(0)}$$ $\displaystyle x^{(2n-k)}=0 \; , \forall k\in N$ $\displaystyle {(xe^x)}^{(2n)}=e^x (2n+x)=2ne^x +xe^x$ $\displaystyle {(xe^x)}^{n}=ne^x+xe^x$ Last edited by idontknow; October 31st, 2017 at 10:42 AM.
 October 31st, 2017, 11:48 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,791 Thanks: 923 I'm not really sure what you are asking but you can easily confirm your final statement using induction. Thanks from topsquark and idontknow
October 31st, 2017, 07:34 PM   #3
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Quote:
 Originally Posted by idontknow Check if this holds true .... (post your check) N-th derivative example without inspection of first , second , third ... derivative $\displaystyle y=xe^x \;$ $$(uv)^{(n)}= \sum \limits_{k=0}^{n} {n \choose k} u^{(n-k}v^{(k)} \;$$ set $\displaystyle u=x , v=e^x$ $$y^{(2n)}= \sum \limits_{k=0}^{2n} {2n \choose k} x^{(2n-k)}(e^x)^{(k)}=e^x \sum \limits_{k=0}^{2n} {2n \choose k} x^{(2n-k)}= {2n \choose 0} x^{(2n)}+{2n \choose 1} x^{(2n-1)}+...+{2n \choose 2n-1} x^{(1)}+{2n \choose 2n} x^{(0)}$$ $\displaystyle x^{(2n-k)}=0 \; , \forall k\in N$ $\displaystyle {(xe^x)}^{(2n)}=e^x (2n+x)=2ne^x +xe^x$ $\displaystyle {(xe^x)}^{n}=ne^x+xe^x$
This is certainly false.

Take $n =2$ and compute this by hand to see the problem.

October 31st, 2017, 07:49 PM   #4
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 Originally Posted by SDK This is certainly false. Take $n =2$ and compute this by hand to see the problem.
I don't know about the derivation but the final line is correct though it should read

$(x e^x)^{(n)} = n e^x + x e^x$

the $n$ is level of differentiation, not exponentiation.

November 1st, 2017, 05:06 AM   #5
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 Originally Posted by romsek I don't know about the derivation but the final line is correct though it should read $(x e^x)^{(n)} = n e^x + x e^x$ the $n$ is level of differentiation, not exponentiation.
I'm referring to the claim that
$(uv)^{(n)} = \sum_{j=0}^n \binom{n}{j}u^{(j)}v^{(n-j)}$

November 1st, 2017, 11:54 AM   #6
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Quote:
 Originally Posted by SDK I'm referring to the claim that $(uv)^{(n)} = \sum_{j=0}^n \binom{n}{j}u^{(j)}v^{(n-j)}$

When n= 2 that formula is $\displaystyle (uv)''= uv''+ 2u'v'+ u''v$.

By the product rule, $\displaystyle (uv)'= u'v+ uv'$ and then $\displaystyle (uv)''= (u'v)'+ (uv')'= (u''v+ u'v')+ (u'v'+ uv'')= u''v+ 2u'v'+ u''v$ so it certainly is true when n= 2.

Last edited by Country Boy; November 1st, 2017 at 11:57 AM.

November 1st, 2017, 04:28 PM   #7
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 Originally Posted by Country Boy And your claim is incorrect. When n= 2 that formula is $\displaystyle (uv)''= uv''+ 2u'v'+ u''v$. By the product rule, $\displaystyle (uv)'= u'v+ uv'$ and then $\displaystyle (uv)''= (u'v)'+ (uv')'= (u''v+ u'v')+ (u'v'+ uv'')= u''v+ 2u'v'+ u''v$ so it certainly is true when n= 2.
I guess I should have worked it out before commenting. Holy crap where has this been my whole life?

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