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watchmath Group TitleBest ResponseYou've already chosen the best response.0
induction on m
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
this is what i have so far \[\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m1}nx^{n1}\\ &=\mathrm D^{m2}n(n1)x^{n2}\\ &=\mathrm D^{m3}n(n1)(n2)x^{n3}\\ &=\qquad\vdots\\ &=\mathrm D^{mn}n(n1)(n2)\dots(nm)x^{nm}\\ &=\mathrm D^{mn}\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{mm}n(n1)(n2)\dots(mm)x^{mm}\\ &=n! \end{align*}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
im not sure if i made a mistake , of if the thing im trying to deduce isn't right
 one year ago

watchmath Group TitleBest ResponseYou've already chosen the best response.0
for m=1 the statement is correct. Assume it is true for m1 which means that \[D^{m1}x^n=\frac{x^{n+1m}n!}{n+1m}\].
 one year ago

watchmath Group TitleBest ResponseYou've already chosen the best response.0
Now \[D^mx^n=D\left(D^{m1}x^n\right)=\frac{(n+1m)x^{nm}n!}{(n+1m)!}=\frac{x^{nm}n!}{(nm)!}\]
 one year ago

watchmath Group TitleBest ResponseYou've already chosen the best response.0
There is a typo\[D^{m1}x^n=\frac{x^{n+1m}n!}{(n+1m)!}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i dont understand
 one year ago

watchmath Group TitleBest ResponseYou've already chosen the best response.0
have you learn mathematical induction?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i dont really like induction,
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
it always confuses me ,
 one year ago

watchmath Group TitleBest ResponseYou've already chosen the best response.0
You'd better learn it again then :D.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
do i have to use induction for this question ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dosent the D^{mn} disappear because n≥m or something like that?
 one year ago

watchmath Group TitleBest ResponseYou've already chosen the best response.0
yours also fine actually. Just remember that D^0(f)=f
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
\(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m1}nx^{n1}\\ &=\mathrm D^{m2}n(n1)x^{n2}\\ &=\mathrm D^{m3}n(n1)(n2)x^{n3}\\ &=\qquad\vdots\\ &=\mathrm D^{mn}n(n1)(n2)\dots\color{blue}{(n(m1))}x^{nm}\\ &=\mathrm D^{mn}\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{mm}n(n1)(n2)\dots\color{blue}{(n(n1))}x^{mm}\\ &=n! \end{align*}\)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
n's are m's
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
whats the doubt now?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\mathrm D^{mn}\frac{n!}{(nm)!}x^{nm}=\frac{n!}{(nm)!}x^{nm}\]?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
\(\begin{align*} &n\in\mathbb N,\qquad m\leq n\\ \\ \mathrm D^mx^n&=\mathrm D^{m1}nx^{n1}\\ &=\mathrm D^{m2}n(n1)x^{n2}\\ &=\mathrm D^{m3}n(n1)(n2)x^{n3}\\ &=\qquad\vdots\\ &=\mathrm D^{m\color{red}{\large m}} n(n1)(n2)\dots\color{blue}{(n(m1))}x^{nm}\\ &=\mathrm D^{mm}\frac{n!}{(nm)!}x^{nm}\\ &=\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ \mathrm D^mx^m&=\mathrm D^{mm}n(n1)(n2)\dots\color{blue}{(n(n1))}x^{mm}\\ &=n! \end{align*}\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
differentiate 'm' time, not n
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
ah that's making some sense now
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
(i got to go right now , ill come back to this question )
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
you got the correction in blue also, right ? (n(m1)).
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[ \begin{align*} &m,n\in\mathbb N,\qquad m\leq n\\ \\ \operatorname D^mx^n&=\operatorname D^{m1}nx^{n1}\\ &=\operatorname D^{m2}n(n1)x^{n2}\\ &=\operatorname D^{m3}n(n1)(n2)x^{n3}\\ &=\operatorname D^{mm}n(n1)(n2)\dots(nm)x^{nm}\\ &=\frac{n!}{(nm)!}x^{nm}\\ &\\ &\\ &\\ \operatorname D^mx^m&=\frac{n!}{(mm)!}x^{mm}\\ &=n!\\ \end{align*} \]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
@hartnn , i'm not sure about the stuff in blue ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
@watchmath , deduction is not induction
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
ah another mistake \[\operatorname D^{\color{lime}n}x^{\color{lime}n}=\frac{n!}{(mm)!}x^{mm} =n!\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oh, I get it.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Isn't that it? You solved your own question.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i kinda want to understand this result a little better, \[\operatorname D^nx^n =n!\], the nth derivative of x to the power n is equal to n factorial
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Use the powerrule repeatedly.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\(\operatorname D^n e^x =e^x\) how does e and ! related
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
You can take the factorial of anything.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Gamma Function bro.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Plus that is not \(x^n\).
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\operatorname D^n(e^x)^n =e^xn!\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Ooooh. The Chain Rule?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\color{red}*\]\[\operatorname D^n(e^x)^n=\operatorname D^n(e^{xn}) =n!e^{xn}\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I don't know what you are trying to say... try the Chain Rule.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
In which question, you have what doubt ? mention clearly, plz.....
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
the blue writing isn't right
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
for m2>n(n(21)) for m3>n(n1)(n(31)) for m4>n(n1)(n2)(n(41)) . . for mm>n(n1)....(n(m1))
 one year ago
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