## UnkleRhaukus 2 years ago Calculate $$\mathrm D^mx^n$$

1. UnkleRhaukus

2. watchmath

induction on m

3. UnkleRhaukus

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

im not sure if i made a mistake , of if the thing im trying to deduce isn't right

5. watchmath

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

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

There is a typo$D^{m-1}x^n=\frac{x^{n+1-m}n!}{(n+1-m)!}$

8. UnkleRhaukus

i dont understand

9. watchmath

have you learn mathematical induction?

10. UnkleRhaukus

i dont really like induction,

11. UnkleRhaukus

it always confuses me ,

12. watchmath

You'd better learn it again then :D.

13. UnkleRhaukus

do i have to use induction for this question ?

14. UnkleRhaukus

dosent the D^{m-n} disappear because n≥m or something like that?

15. watchmath

yours also fine actually. Just remember that D^0(f)=f

16. UnkleRhaukus

hmmm

17. hartnn

\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

n's are m's

19. hartnn

whats the doubt now?

20. UnkleRhaukus

$\mathrm D^{m-n}\frac{n!}{(n-m)!}x^{n-m}=\frac{n!}{(n-m)!}x^{n-m}$?

21. hartnn

\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

differentiate 'm' time, not n

23. hartnn

*times.

24. UnkleRhaukus

ah that's making some sense now

25. UnkleRhaukus

(i got to go right now , ill come back to this question )

26. hartnn

you got the correction in blue also, right ? (n-(m-1)).

27. UnkleRhaukus

\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

@hartnn , i'm not sure about the stuff in blue ?

29. UnkleRhaukus

@watchmath , deduction is not induction

30. UnkleRhaukus

ah another mistake $\operatorname D^{\color{lime}n}x^{\color{lime}n}=\frac{n!}{(m-m)!}x^{m-m} =n!$

31. ParthKohli

Oh, I get it.

32. UnkleRhaukus

good

33. ParthKohli

Isn't that it? You solved your own question.

34. UnkleRhaukus

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

Use the power-rule repeatedly.

36. UnkleRhaukus

$$\operatorname D^n e^x =e^x$$ how does e and ! related

37. ParthKohli

Why?

38. ParthKohli

You can take the factorial of anything.

39. ParthKohli

Gamma Function bro.

40. UnkleRhaukus

yeah

41. ParthKohli

Plus that is not $$x^n$$.

42. UnkleRhaukus

$\operatorname D^n(e^x)^n =e^xn!$

43. ParthKohli

Ooooh. The Chain Rule?

44. UnkleRhaukus

$\color{red}*$$\operatorname D^n(e^x)^n=\operatorname D^n(e^{xn}) =n!e^{xn}$

45. ParthKohli

I don't know what you are trying to say... try the Chain Rule.

46. hartnn

In which question, you have what doubt ? mention clearly, plz.....

47. UnkleRhaukus

the blue writing isn't right

48. hartnn

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))