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UnkleRhaukus

  • one year ago

\[\mathrm D^6\big(x^3e^{ax}\big)\]

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  1. UnkleRhaukus
    • one year ago
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    \[\begin{equation*}\mathrm D \equiv \frac{\mathrm d}{\mathrm dx}\end{equation*} \]

  2. UnkleRhaukus
    • one year ago
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    \[\begin{align*} \mathrm D^6\big(x^3e^{ax}\big)&\\ &=\mathrm D^5\big(3x^2+ax^3\big)e^{ax}\\ &=\mathrm D^4\big(6x+3ax^2+3ax^2+a^2x^3\big)e^{ax}\\ &=\mathrm D^4\big(6x+6ax^2+a^2x^3\big)e^{ax}\\ &=\mathrm D^3\big(6+12ax+3a^2x^2+6ax+6a^2x^2+a^3x^3\big)e^{ax}\\ &=\mathrm D^3\big(6+18ax+9a^2x^2+a^3x^3\big)e^{ax}\\ &=\mathrm D^2\big(18a+18a^2x+3a^3x^2+6a+18a^2x+9a^3x^2+a^4x^3\big)e^{ax}\\ &=\mathrm D^2\big(24a+36a^2x+12a^3x^2+a^4x^3\big)e^{ax}\\ &=\mathrm D\big(36a^2+24a^3x+3a^4x^2+24a^2+36a^3x+12a^4x^2+a^5x^3\big)e^{ax}\\ &=\mathrm D\big(60a^2+60a^3x+15a^4x^2+a^5x^3\big)e^{ax}\\ &=\big(60a^3+30a^4x+3a^5x^2+60a^3+60a^4x+15a^5x^2+a^6x^3\big)e^{ax}\\ &=\big(120a^3+90a^4x+18a^5x^2+a^6x^3\big)e^{ax}\\ &=\big(120+90ax+18a^2x^2+a^3x^3\big)a^3e^{ax}\\ \end{align*}\]

  3. UnkleRhaukus
    • one year ago
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    Is there an easier way to do this ?

  4. hartnn
    • one year ago
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    D is derivative operator ? can't u use general formula for \(D^n(x^me^{at})\)

  5. hartnn
    • one year ago
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    *ax

  6. UnkleRhaukus
    • one year ago
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    \[\mathrm D^n(x^me^{ax})=?\]

  7. hartnn
    • one year ago
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    i thought there was general formula....but apparently there isn't xD

  8. UnkleRhaukus
    • one year ago
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    some thing with factorials !

  9. hartnn
    • one year ago
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    tried Leibnitz theorem ? \(D^n(uv)\)

  10. UnkleRhaukus
    • one year ago
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    what is that?

  11. UnkleRhaukus
    • one year ago
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    oh

  12. hartnn
    • one year ago
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    does that become easier ? i doubt :P

  13. sirm3d
    • one year ago
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    \[ D^6(x^3 e^{ax})=e^{ax}D^6(x^3)+6D^5(x^3)D(e^{ax}) +\cdots+6D(x^3)D^5(e^{ax})+x^3D^6(e^{ax})\]\[D^6x^3=0, \;D^5x^3=0,\; D^4x^3=0,\;D^3x^3=3!\]

  14. sirm3d
    • one year ago
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    \[D^2x^3=6x,\;Dx^3=3x^2,\;D^n(e^{ax})=a^ne^{ax}\]\[D^6(x^3e^{ax})=20(3!a^3e^{ax})+15(6xa^4e^{ax})+6(3x^2a^5e^{ax})+(x^3a^6e^{ax})\]it appears the solution is shorter

  15. experimentX
    • one year ago
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    |dw:1357054140625:dw| i guess Leibniz theorem is better.

  16. UnkleRhaukus
    • one year ago
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    \[\cdot\]

  17. mukushla
    • one year ago
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    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

  18. mukushla
    • one year ago
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    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 !!!!

  19. mukushla
    • one year ago
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    btw maybe that will be a clue for someone :)

  20. mukushla
    • one year ago
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    also @TuringTest will like to try this

  21. UnkleRhaukus
    • one year ago
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    \[\boxed{\frac{\mathrm d^n}{\mathrm dx^n}(f\cdot g)=\sum\limits_{k=0}^n\frac{n!}{k!(n-k)!}\cdot\frac{\mathrm d^kf}{\mathrm dx^k}\cdot\frac{\mathrm d^{n-k}g}{\mathrm dx^{n-k}}}\] \[\begin{align*} \mathrm D^6\big(x^3e^{ax}\big)&=\sum\limits_{k=0}^6\frac{6!}{k!(6-k)!}\cdot\frac{\mathrm d^k}{\mathrm dx^k}(x^3)\cdot\frac{\mathrm d^{6-k}}{\mathrm dx^{6-k}}(e^{ax})\\ &=6!\left(\frac{x^3a^6}{6!}+\frac{3x^2a^5}{5!}+\frac{6xa^4}{2!4!}+\frac{6a^3}{3!^2}+0+0+0\right)e^{ax}\\ %&=\left(x^3a^6+18x^2a^5+90xa^4+120a^3\right)e^{ax}\\ &=\left(x^3a^3+18x^2a^2+90xa+120\right)a^3e^{ax}\\ \end{align*}\] thankyou @hartnn @sirm3d @experimentX @mukushla

  22. UnkleRhaukus
    • one year ago
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    the formula reminds me of convolution

  23. UnkleRhaukus
    • one year ago
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    \[\begin{align*} \mathrm D^4\big(x^2\cos(2x)\big)\\ &=\sum\limits_{k=0}^4\frac{4!}{k!(4-k)!}\cdot\frac{\mathrm d^k(x^2)}{\mathrm dx^k}\cdot\frac{\mathrm d^{4-k}(\cos(2x))}{\mathrm dx^{4-k}}\\ &=4!\left(\frac{x^2\times16\cos(2x)}{4!}+\frac{2x\times8\sin(2x)}{3!}+\frac{2\times-4\cos(2x)}{2!^2}+0+0\right)\\ &=16x^2\cos(2x)+64x\sin(2x)-48\cos(2x) \end{align*}\]

  24. UnkleRhaukus
    • one year ago
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    \[\begin{align*} &\mathrm D^4\big(e^{2x}\cos(3x)\big)&\\ &=\sum\limits_{k=0}^4\frac{4!}{k!(4-k)!}\cdot\frac{\mathrm d^k(e^{2x})}{\mathrm dx^k}\cdot\frac{\mathrm d^{4-k}(\cos(3x))}{\mathrm dx^{4-k}}\\ &=4!\left(\frac{e^{2x}\times81\cos(3x)}{4!}+\frac{2e^{2x}\times27\sin(3x)}{3!}+\frac{4e^{2x}\times-9\cos(3x)}{2!^2}\right.\\ &\qquad\qquad\qquad\qquad\left.+\frac{8e^{2x}\times-3\sin(3x)}{3!}+\frac{16e^{2x}\cos(3x)}{4!}\right)\\ &=81e^{2x}\cos(3x)+216e^{2x}\sin(3x)-216e^{2x}\cos(3x)-96e^{2x}\sin(3x)+16e^{2x}\cos(3x)\\ &=120e^{2x}\sin(3x)-119e^{2x}\cos(3x) \end{align*}\] woo!

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