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ganeshie8
 one year ago
show that
\[\int x^ne^x\,dx = e^x\left[x^nnx^{n1}+n(n1)x^{n2}n(n1)(n2)x^{n3}\\~~~~~~~~~+\cdots + (1)^nn!\right] \]
ganeshie8
 one year ago
show that \[\int x^ne^x\,dx = e^x\left[x^nnx^{n1}+n(n1)x^{n2}n(n1)(n2)x^{n3}\\~~~~~~~~~+\cdots + (1)^nn!\right] \]

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0example : http://www.wolframalpha.com/input/?i=%5Cint+x%5E5e%5Ex

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.3\[I_n=\int\limits x^n e^xdx =\\x^n e^x \int\limits e^x (nx^{n1})dx =\\ \]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.3\[I_n=x^n*e^x n*I_{n1}\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.3\[I_{n}=x^n*e^x n*(x^{n1}*e^x (n1)I_{n2})\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.3and so on ... to get formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Ahh IBP works out smoothly!! Any alternatives ? This feels like a famous integral, I'm pretty sure there must be many ways to work this...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0The numbers happen to be coming from the columns of below inverse matrix http://www.wolframalpha.com/input/?i=inverse+%7B%7B1%2C1%2C0%2C0%2C0%2C0%7D%2C%7B0%2C1%2C2%2C0%2C0%2C0%7D%2C%7B0%2C0%2C1%2C3%2C0%2C0%7D%2C%7B0%2C0%2C0%2C1%2C4%2C0%7D%2C%7B0%2C0%2C0%2C0%2C1%2C5%7D%2C%7B0%2C0%2C0%2C0%2C0%2C1%7D%7D for example, the last column gives the coeffieitnes for \(\int x^5 e^x\,dx\) and the fifth column gives the coefficients for \(\int x^4e^x\, dx\) etc... makes me wonder if there is a direct formula to work that inverse, that matrix itself looks pretty patterny...

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2\[ \begin{align*} R&=e^x\left[x^nnx^{n1}+n(n1)n^{x2}n(n1)(n2)x^{n3}+\cdots+(1)^nn!\right]\\ &=e^x\sum_{k=0}^n\frac{(1)^k n!}{(nk)!}x^{(nk)}\\ R'&=e^x\sum_{k=0}^n\frac{(1)^k n!}{(nk)!}x^{(nk)}+e^x\sum_{k=0}^{n1}\frac{(1)^k n!}{(n1k)!}x^{(nk1)}\\ R'&=e^x\sum_{k=0}^n\frac{(1)^k n!}{(nk)!}x^{(nk)}+e^x\sum_{k=1}^{n}\frac{(1)^{k+1} n!}{(nk)!}x^{(nk)}\\ &=e^xx^n \end{align*} \] The idea is correct but the index shift in sum is probably not.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I thought you would take left hand side and replace \(e^x\) by its series. But your method looks interesting!

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2Just find a closed form sum of the right hand side and differentiate both sides.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Here this looks less painful: \[\int x^ne^x dx = \int x^n \sum_{k=0}^\infty \frac{x^k}{k!} dx = \int \sum_{k=0}^\infty \frac{x^{k+n}}{k!} dx = \sum_{k=0}^\infty \int \frac{x^{k+n}}{k!} dx \] \[ \sum_{k=0}^\infty \int \frac{x^{k+n}}{k!} dx = \sum_{k=0}^\infty \frac{x^{k+n+1}}{k!(k+n+1)} + C \] \[= x^{n+1} \sum_{k=0}^\infty \frac{x^{k}}{(k+1)!+n k!} + C \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0is it easy to get a finite sum out of that, looks that has to break up as some weird combinations of \(e^x\) and polynomials

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah I'm sorta working on breaking that up using partial fractions right now instead of factoring it like I did. This is a fun problem I tried about 3 years ago but sorta gave up on it. It's sorta related to the gamma function I think, it's pretty interesting, what makes you interested in it? :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0linear transformations ofcourse! i remember something similar from one of your posts : \[\int e^{ax}\sin(bx) \,dx \] If I recall correctly, you had worked it by choosing basis as \(\{e^{ax}\cos(bx),~e^{ax}\sin(bx)\}\) or something similar to that

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ahhh yeah that's a fun one. :P

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I think specifically when I was showing mathslover that XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah right! that post

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0you guys have elephant memory :\ i cant even remember what food i had for dinner xD

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Hahaha ok but that was a fun thread though so that's why it stands out so much, even if it was like 1 or 2 years ago XD

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2No I am not expanding \(e^x\) but rewriting the polynomial on the right hand side using sigma notation and differentiate term by term.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2\[ \left[x^nnx^{n1}+n(n1)n^{x2}n(n1)(n2)x^{n3}+\cdots+(1)^nn!\right]\\ =\sum_{k=0}^n\frac{(1)^k n!}{(nk)!}x^{(nk)}\\ \]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ok here's a possible linear transformation way, I think it'll work: So I'll just write the basis in two different ways: B={\(e^x,xe^x,x^2e^x,x^3e^x,\dots \)} = {\(y_0,y_1,y_2,y_3, \dots\)} \[Dy_n = y_n+ny_{n1}\] So you can try that out, and then see we have two matrices sorta working here, the first is an identity matrix and the second is sorta like a shift matrix and has n in it. This is actually just the regular derivative matrix for if our basis didn't have \(e^x\) in there (you can kinda believe this from the product rule just multiplying \(e^x\) and then differentiating the polynomial): \[Dy_n = I y_n+Ay_n\] So to be clear about what I'm writing: \[Iy_n = y_n\]\[Ay_n = n y_{n1}\] \[D=I+A\] Ok so the problem we have is that \(A\) is an infinite matrix. Luckily though, we don't need to know all the entries, just enough up to what we're interested in, since we can see that the derivative matrix will only map from our basis at the place \(n\) to \(n1\) that means the inverse matrix which exactly reverses this can at most map backwards to \(n\) and \(n+1\). So that means we just need to find the derivative matrix up to \(n+1\) more than what we want to integrate, simple as that. I'll show an example. \[\int x^2e^x dx\] So we have to calculate the derivative matrix for up to 3. So that means we have: \[D = I+ A = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0& 1 & 0 & 0\\ 0 & 0& 2& 0\\ 0 & 0 & 0 & 3\\ 0 & 0 & 0 & 0 \end{bmatrix} =\begin{bmatrix} 1& 1 & 0 & 0\\ 0 & 1& 2& 0\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 1 \end{bmatrix} \] Now we invert this matrix... with matlab... \[ D^{1} = \begin{bmatrix} 1& 1 &2 & 6 \\ 0 & 1& 2& 6\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 1 \end{bmatrix} \] Multiply by the vector representing our integral: \[\begin{bmatrix} 1& 1 &2 & 6 \\ 0 & 1& 2& 6\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix} = \begin{bmatrix} 2\\ 2\\ 1\\ 0 \end{bmatrix}\] Ok moment of truth, hopefully I didn't mess this up... \[\int x^2e^xdx = e^x(22x+x^2) + C\] Hey it checks out :P Also, maybe we can possibly find a general form for the inverse of this matrix?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Looking through this it appears that maybe my reasoning was off, since if we throw in the vector corresponding to \(\int x^3e^xdx\) it seems to work as well. So perhaps for that last example we could have done it with a 3x3 matrix.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2\[ B=\{x^ne^x,x^{n1}e^x,x^{n2}e^x,\dots,xe^x,e^x\}=\{y_1,y_2,y_3,\dots,y_{n+1}\}\\ D_{n}y_n=y_n+ny_{n1}\\ D_{n}= \begin{bmatrix} 1 &0&\cdots&\cdots&\cdots&\cdots&0\\ n&1&\ddots &\ddots&\ddots&\ddots&\vdots\\ 0&n1&1&\ddots &\ddots&\ddots&\vdots\\ \vdots&0&n2&1&\ddots &\ddots&\vdots\\ \vdots&\vdots&0&\ddots &\ddots&\ddots &\vdots \\ \vdots&\vdots &\vdots&\ddots&2&1&0 \\ 0&0&0&0&0&1&1 \end{bmatrix}\;(n+1)\times (n+1)\text{ matrix}\\ D_n^{1}\text{ exists as }\det(D_n)=1 \text{ as }D_n\text{ is lower triangular.}\\ \text{Moreover, }D_n^{1}\text{ is an integer matrix as }\det(D_n)=1\text{ (1 is invertible in }\mathbb{Z}\text{)} \]
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