Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

JonaskBest ResponseYou've already chosen the best response.0
\[(fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)f^{k}g^{nk}\] where\[f^{(n)}=\frac{ d^{(n)} f}{ dx^{n} }\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
I am stuck on \[\left(\begin{matrix}m \\ k1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=?\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
http://physicspages.com/2011/03/22/generalizedproductruleleibnizsformula/
 one year ago

Hares333Best ResponseYou've already chosen the best response.1
http://www.math.ucsd.edu/~wgarner/math20a/prodrule.htm
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
\[\left(\begin{matrix}m \\ k1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=\frac{ m! }{ (k1)!(mk+1)! }+\frac{ m! }{ k!(mk!) }\]\[=\frac{ m! }{ (k1)!(mk)! }\left[ \frac{ 1 }{ mk+1 }+\frac{ 1 }{ k } \right]\]\[=\frac{ m! }{ (k1)!(mk)! }\left[ \frac{ k+mk+1 }{ k(mk+1) } \right]\]\[=\frac{ m! }{ (k1)!(mk)! }.\frac{ m+1 }{ k(mk+1) }\]\[=\frac{ (m+1)! }{ k!(mk+1)! }=\left(\begin{matrix}m+1 \\ k\end{matrix}\right)\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
@Jonask hope this helped
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
thanks guys ,appreciated
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.