UnkleRhaukus
  • UnkleRhaukus
\[\mathrm D^6\big(x^3e^{ax}\big)\]
Mathematics
katieb
  • katieb
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katieb
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UnkleRhaukus
  • UnkleRhaukus
\[\begin{equation*}\mathrm D \equiv \frac{\mathrm d}{\mathrm dx}\end{equation*} \]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} \mathrm D^6\big(x^3e^{ax}\big)&\\ &=\mathrm D^5\big(3x^2+ax^3\big)e^{ax}\\ &=\mathrm D^4\big(6x+3ax^2+3ax^2+a^2x^3\big)e^{ax}\\ &=\mathrm D^4\big(6x+6ax^2+a^2x^3\big)e^{ax}\\ &=\mathrm D^3\big(6+12ax+3a^2x^2+6ax+6a^2x^2+a^3x^3\big)e^{ax}\\ &=\mathrm D^3\big(6+18ax+9a^2x^2+a^3x^3\big)e^{ax}\\ &=\mathrm D^2\big(18a+18a^2x+3a^3x^2+6a+18a^2x+9a^3x^2+a^4x^3\big)e^{ax}\\ &=\mathrm D^2\big(24a+36a^2x+12a^3x^2+a^4x^3\big)e^{ax}\\ &=\mathrm D\big(36a^2+24a^3x+3a^4x^2+24a^2+36a^3x+12a^4x^2+a^5x^3\big)e^{ax}\\ &=\mathrm D\big(60a^2+60a^3x+15a^4x^2+a^5x^3\big)e^{ax}\\ &=\big(60a^3+30a^4x+3a^5x^2+60a^3+60a^4x+15a^5x^2+a^6x^3\big)e^{ax}\\ &=\big(120a^3+90a^4x+18a^5x^2+a^6x^3\big)e^{ax}\\ &=\big(120+90ax+18a^2x^2+a^3x^3\big)a^3e^{ax}\\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
Is there an easier way to do this ?

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hartnn
  • hartnn
D is derivative operator ? can't u use general formula for \(D^n(x^me^{at})\)
hartnn
  • hartnn
*ax
UnkleRhaukus
  • UnkleRhaukus
\[\mathrm D^n(x^me^{ax})=?\]
hartnn
  • hartnn
i thought there was general formula....but apparently there isn't xD
UnkleRhaukus
  • UnkleRhaukus
some thing with factorials !
hartnn
  • hartnn
tried Leibnitz theorem ? \(D^n(uv)\)
UnkleRhaukus
  • UnkleRhaukus
what is that?
hartnn
  • hartnn
https://docs.google.com/viewer?a=v&q=cache:9WaUs5ovMl8J:www.math.osu.edu/~nevai.1/H16x/DOCUMENTS/leibniz_product_formula_H6.pdf+&hl=en&gl=in&pid=bl&srcid=ADGEEShrDfDyJl8y40cSZzEHIU0yJcWw6JunzRb9a0fIS20lbReliJzWWroyHm2gc_YX4UUZ5J_bQaXUfaa1e80MB67e9UYKKMnKVVYQZmrG0G4oWkzj1wshx_2H4Pbh6HtYYhruMq2y&sig=AHIEtbTOMTfpYiFiWbXziBk9x2R7lGVEfA
UnkleRhaukus
  • UnkleRhaukus
oh
hartnn
  • hartnn
does that become easier ? i doubt :P
sirm3d
  • sirm3d
\[ D^6(x^3 e^{ax})=e^{ax}D^6(x^3)+6D^5(x^3)D(e^{ax}) +\cdots+6D(x^3)D^5(e^{ax})+x^3D^6(e^{ax})\]\[D^6x^3=0, \;D^5x^3=0,\; D^4x^3=0,\;D^3x^3=3!\]
sirm3d
  • sirm3d
\[D^2x^3=6x,\;Dx^3=3x^2,\;D^n(e^{ax})=a^ne^{ax}\]\[D^6(x^3e^{ax})=20(3!a^3e^{ax})+15(6xa^4e^{ax})+6(3x^2a^5e^{ax})+(x^3a^6e^{ax})\]it appears the solution is shorter
experimentX
  • experimentX
|dw:1357054140625:dw| i guess Leibniz theorem is better.
UnkleRhaukus
  • UnkleRhaukus
\[\cdot\]
anonymous
  • anonymous
i miss beautiful problems like this these days :( lets see what we'll get with use of Leibniz\[(uv)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)u^{(n-k)}v^k\]\[u=x^m\]\[v=e^{ax}\]so we have\[(x^m e^{ax})^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)(x^m)^{(n-k)}(e^{ax})^k=e^{ax} \sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)(x^m)^{(n-k)}(a)^k\]there is a little problem here
anonymous
  • anonymous
if \(m=n-k\) \[(x^m)^{(n-k)}=m!\]if \(m>n-k\)\[(x^m)^{(n-k)}=m(m-1)(m-2)...(m-n+k+1)x^{n-k} \]ohhh this is hard to get !!!!
anonymous
  • anonymous
btw maybe that will be a clue for someone :)
anonymous
  • anonymous
also @TuringTest will like to try this
UnkleRhaukus
  • UnkleRhaukus
\[\boxed{\frac{\mathrm d^n}{\mathrm dx^n}(f\cdot g)=\sum\limits_{k=0}^n\frac{n!}{k!(n-k)!}\cdot\frac{\mathrm d^kf}{\mathrm dx^k}\cdot\frac{\mathrm d^{n-k}g}{\mathrm dx^{n-k}}}\] \[\begin{align*} \mathrm D^6\big(x^3e^{ax}\big)&=\sum\limits_{k=0}^6\frac{6!}{k!(6-k)!}\cdot\frac{\mathrm d^k}{\mathrm dx^k}(x^3)\cdot\frac{\mathrm d^{6-k}}{\mathrm dx^{6-k}}(e^{ax})\\ &=6!\left(\frac{x^3a^6}{6!}+\frac{3x^2a^5}{5!}+\frac{6xa^4}{2!4!}+\frac{6a^3}{3!^2}+0+0+0\right)e^{ax}\\ %&=\left(x^3a^6+18x^2a^5+90xa^4+120a^3\right)e^{ax}\\ &=\left(x^3a^3+18x^2a^2+90xa+120\right)a^3e^{ax}\\ \end{align*}\] thankyou @hartnn @sirm3d @experimentX @mukushla
UnkleRhaukus
  • UnkleRhaukus
the formula reminds me of convolution
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} \mathrm D^4\big(x^2\cos(2x)\big)\\ &=\sum\limits_{k=0}^4\frac{4!}{k!(4-k)!}\cdot\frac{\mathrm d^k(x^2)}{\mathrm dx^k}\cdot\frac{\mathrm d^{4-k}(\cos(2x))}{\mathrm dx^{4-k}}\\ &=4!\left(\frac{x^2\times16\cos(2x)}{4!}+\frac{2x\times8\sin(2x)}{3!}+\frac{2\times-4\cos(2x)}{2!^2}+0+0\right)\\ &=16x^2\cos(2x)+64x\sin(2x)-48\cos(2x) \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} &\mathrm D^4\big(e^{2x}\cos(3x)\big)&\\ &=\sum\limits_{k=0}^4\frac{4!}{k!(4-k)!}\cdot\frac{\mathrm d^k(e^{2x})}{\mathrm dx^k}\cdot\frac{\mathrm d^{4-k}(\cos(3x))}{\mathrm dx^{4-k}}\\ &=4!\left(\frac{e^{2x}\times81\cos(3x)}{4!}+\frac{2e^{2x}\times27\sin(3x)}{3!}+\frac{4e^{2x}\times-9\cos(3x)}{2!^2}\right.\\ &\qquad\qquad\qquad\qquad\left.+\frac{8e^{2x}\times-3\sin(3x)}{3!}+\frac{16e^{2x}\cos(3x)}{4!}\right)\\ &=81e^{2x}\cos(3x)+216e^{2x}\sin(3x)-216e^{2x}\cos(3x)-96e^{2x}\sin(3x)+16e^{2x}\cos(3x)\\ &=120e^{2x}\sin(3x)-119e^{2x}\cos(3x) \end{align*}\] woo!

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