i hv jst started with leibnitz`s theorem and cant understand it plz someone explane me this one

Question

anonymous · Answer

if $$y= x ^{2}e ^{x}$$
then show that $$y _{n} = y _{2}\frac{ n(n-1) }{ 2 } - n(n-2) y _{2} + \frac{ (n-1)(n-2) }{ 2 }y$$

amistre64 · Answer

we might want to start with defining what you think the leibniz thrm says ...

anonymous · Answer

i know the formula and its method i hv done just a few questions which dealt with just expanding or solving them by applying the formula this is the first of show that kind...
and i am not getting the answer

amistre64 · Answer

this one?
$$\sum_{k=0}^{n}~\binom{n}{k}\frac{d^k}{dx^k}[u(x)]~\frac{d^{n-k}}{dx^{n-k}}[v(x)]$$

amistre64 · Answer

or do we have to brute it out to find a recursive pattern that would end up being that?

amistre64 · Answer

$$x^2e^x$$$$2xe^x+x^2e^x$$$$2e^x+4xe^x+x^2e^x$$$$6e^x+6xe^x+x^2e^x$$$$12e^x+8xe^x+x^2e^x$$
etc...

anonymous · Answer

yes that was of this kind yes

amistre64 · Answer

so successive product rules ...

amistre64 · Answer

$$f = 0e^x+0xe^x+x^2e^x$$
$$f^1 = 0e^x+2xe^x+x^2e^x$$
$$f^2 = 2e^x+4xe^x+x^2e^x$$
$$f^3 = 6e^x+6xe^x+x^2e^x$$
$$f^4 = 12e^x+8xe^x+x^2e^x$$
$$f^5 = 20e^x+10xe^x+x^2e^x$$
... it appears to be this pattern to me
$$f^n=n(n-1)e^x+2n~xe^x+x^2e^x$$

amistre64 · Answer

can we conform that to the setup you have?

amistre64 · Answer

looks like someone else came to the same conclusion ... Example 7.4.2 is similar except a=1 in your case http://wdjoyner.com/teach/calc1-sage/html/node106.html