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UnkleRhaukus

  • one year ago

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

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

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

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

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

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

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

  8. UnkleRhaukus
    • one year ago
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    i dont understand

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

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

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

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

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

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

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

  16. UnkleRhaukus
    • one year ago
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    hmmm

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

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

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

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

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

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

  23. hartnn
    • one year ago
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    *times.

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

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

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

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

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

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

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

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

  32. UnkleRhaukus
    • one year ago
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    good

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

  34. UnkleRhaukus
    • one year ago
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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

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

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

  37. ParthKohli
    • one year ago
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    Why?

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

  39. ParthKohli
    • one year ago
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    Gamma Function bro.

  40. UnkleRhaukus
    • one year ago
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    yeah

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

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

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

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

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

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

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

  48. hartnn
    • one year ago
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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))

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