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October 31st, 2017, 11:40 AM   #1
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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 11:42 AM.
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October 31st, 2017, 12:48 PM   #2
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I'm not really sure what you are asking but you can easily confirm your final statement using induction.
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October 31st, 2017, 08:34 PM   #3
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Quote:
Originally Posted by idontknow View Post
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.
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October 31st, 2017, 08:49 PM   #4
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Quote:
Originally Posted by SDK View Post
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.
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November 1st, 2017, 06:06 AM   #5
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Originally Posted by romsek View Post
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
\[(uv)^{(n)} = \sum_{j=0}^n \binom{n}{j}u^{(j)}v^{(n-j)}\]
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November 1st, 2017, 12:54 PM   #6
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Quote:
Originally Posted by SDK View Post
I'm referring to the claim that
\[(uv)^{(n)} = \sum_{j=0}^n \binom{n}{j}u^{(j)}v^{(n-j)}\]
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.
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Last edited by Country Boy; November 1st, 2017 at 12:57 PM.
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November 1st, 2017, 05:28 PM   #7
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Originally Posted by Country Boy View Post
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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