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UnkleRhaukus

Calculate \(\mathrm D^mx^n\)

  • one year ago
  • one year ago

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  1. UnkleRhaukus
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    • one year ago
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  2. watchmath
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    induction on m

    • one year ago
  3. UnkleRhaukus
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    this is what i have so far \[\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\ &=\mathrm D^{m-2}n(n-1)x^{n-2}\\ &=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\qquad\vdots\\ &=\mathrm D^{m-n}n(n-1)(n-2)\dots(n-m)x^{n-m}\\ &=\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots(m-m)x^{m-m}\\ &=n! \end{align*}\]

    • one year ago
  4. UnkleRhaukus
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    im not sure if i made a mistake , of if the thing im trying to deduce isn't right

    • one year ago
  5. watchmath
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    for m=1 the statement is correct. Assume it is true for m-1 which means that \[D^{m-1}x^n=\frac{x^{n+1-m}n!}{n+1-m}\].

    • one year ago
  6. watchmath
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    Now \[D^mx^n=D\left(D^{m-1}x^n\right)=\frac{(n+1-m)x^{n-m}n!}{(n+1-m)!}=\frac{x^{n-m}n!}{(n-m)!}\]

    • one year ago
  7. watchmath
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    There is a typo\[D^{m-1}x^n=\frac{x^{n+1-m}n!}{(n+1-m)!}\]

    • one year ago
  8. UnkleRhaukus
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    i dont understand

    • one year ago
  9. watchmath
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    have you learn mathematical induction?

    • one year ago
  10. UnkleRhaukus
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    i dont really like induction,

    • one year ago
  11. UnkleRhaukus
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    it always confuses me ,

    • one year ago
  12. watchmath
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    You'd better learn it again then :D.

    • one year ago
  13. UnkleRhaukus
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    do i have to use induction for this question ?

    • one year ago
  14. UnkleRhaukus
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    dosent the D^{m-n} disappear because n≥m or something like that?

    • one year ago
  15. watchmath
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    yours also fine actually. Just remember that D^0(f)=f

    • one year ago
  16. UnkleRhaukus
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    hmmm

    • one year ago
  17. hartnn
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    \(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\ &=\mathrm D^{m-2}n(n-1)x^{n-2}\\ &=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\qquad\vdots\\ &=\mathrm D^{m-n}n(n-1)(n-2)\dots\color{blue}{(n-(m-1))}x^{n-m}\\ &=\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots\color{blue}{(n-(n-1))}x^{m-m}\\ &=n! \end{align*}\)

    • one year ago
  18. UnkleRhaukus
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    n's are m's

    • one year ago
  19. hartnn
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    whats the doubt now?

    • one year ago
  20. UnkleRhaukus
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    \[\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}=\frac{n!}{(n-m)!}x^{n-m}\]?

    • one year ago
  21. hartnn
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    \(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m-1}nx^{n-1}\\ &=\mathrm D^{m-2}n(n-1)x^{n-2}\\ &=\mathrm D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\qquad\vdots\\ &=\mathrm D^{m-\color{red}{\large m}} n(n-1)(n-2)\dots\color{blue}{(n-(m-1))}x^{n-m}\\ &=\mathrm D^{m-m}\frac{n!}{(n-m)!}x^{n-m}\\ &=\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{m-m}n(n-1)(n-2)\dots\color{blue}{(n-(n-1))}x^{m-m}\\ &=n! \end{align*}\)

    • one year ago
  22. hartnn
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    differentiate 'm' time, not n

    • one year ago
  23. hartnn
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    *times.

    • one year ago
  24. UnkleRhaukus
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    ah that's making some sense now

    • one year ago
  25. UnkleRhaukus
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    (i got to go right now , ill come back to this question )

    • one year ago
  26. hartnn
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    you got the correction in blue also, right ? (n-(m-1)).

    • one year ago
  27. UnkleRhaukus
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    \[ \begin{align*} &m,n\in\mathbb N,\qquad m\leq n\\ \\ \operatorname D^mx^n&=\operatorname D^{m-1}nx^{n-1}\\ &=\operatorname D^{m-2}n(n-1)x^{n-2}\\ &=\operatorname D^{m-3}n(n-1)(n-2)x^{n-3}\\ &=\operatorname D^{m-m}n(n-1)(n-2)\dots(n-m)x^{n-m}\\ &=\frac{n!}{(n-m)!}x^{n-m}\\ &\\ &\\ &\\ \operatorname D^mx^m&=\frac{n!}{(m-m)!}x^{m-m}\\ &=n!\\ \end{align*} \]

    • one year ago
  28. UnkleRhaukus
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    @hartnn , i'm not sure about the stuff in blue ?

    • one year ago
  29. UnkleRhaukus
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    @watchmath , deduction is not induction

    • one year ago
  30. UnkleRhaukus
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    ah another mistake \[\operatorname D^{\color{lime}n}x^{\color{lime}n}=\frac{n!}{(m-m)!}x^{m-m} =n!\]

    • one year ago
  31. ParthKohli
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    Oh, I get it.

    • one year ago
  32. UnkleRhaukus
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    good

    • one year ago
  33. ParthKohli
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    Isn't that it? You solved your own question.

    • one year ago
  34. UnkleRhaukus
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    i kinda want to understand this result a little better, \[\operatorname D^nx^n =n!\], the n-th derivative of x to the power n is equal to n factorial

    • one year ago
  35. ParthKohli
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    Use the power-rule repeatedly.

    • one year ago
  36. UnkleRhaukus
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    \(\operatorname D^n e^x =e^x\) how does e and ! related

    • one year ago
  37. ParthKohli
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    Why?

    • one year ago
  38. ParthKohli
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    You can take the factorial of anything.

    • one year ago
  39. ParthKohli
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    Gamma Function bro.

    • one year ago
  40. UnkleRhaukus
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    yeah

    • one year ago
  41. ParthKohli
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    Plus that is not \(x^n\).

    • one year ago
  42. UnkleRhaukus
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    \[\operatorname D^n(e^x)^n =e^xn!\]

    • one year ago
  43. ParthKohli
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    Ooooh. The Chain Rule?

    • one year ago
  44. UnkleRhaukus
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    \[\color{red}*\]\[\operatorname D^n(e^x)^n=\operatorname D^n(e^{xn}) =n!e^{xn}\]

    • one year ago
  45. ParthKohli
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    I don't know what you are trying to say... try the Chain Rule.

    • one year ago
  46. hartnn
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    In which question, you have what doubt ? mention clearly, plz.....

    • one year ago
  47. UnkleRhaukus
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    the blue writing isn't right

    • one year ago
  48. hartnn
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    for m-2--->n(n-(2-1)) for m-3--->n(n-1)(n-(3-1)) for m-4--->n(n-1)(n-2)(n-(4-1)) . . for m-m--->n(n-1)....(n-(m-1))

    • one year ago
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