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anonymous
 4 years ago
Can anyone help me with a notation question? In Problem Set 1, 1G5 part b, there is (n over 1) etc with no division line, it's been a long time since I took Calculus in the first place and I can't recall what the significance is? When I look at the answers I'm getting it seems like it should just be n, but their solution for y to (p+q) power throws me for a loop. Any help would be appreciated. Thanks!
anonymous
 4 years ago
Can anyone help me with a notation question? In Problem Set 1, 1G5 part b, there is (n over 1) etc with no division line, it's been a long time since I took Calculus in the first place and I can't recall what the significance is? When I look at the answers I'm getting it seems like it should just be n, but their solution for y to (p+q) power throws me for a loop. Any help would be appreciated. Thanks!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you have a link to the problem set or can you post the problem?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I assume, this is the link... http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/assignments/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't find the problem...please post.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partadefinitionandbasicrules/problemset1/ It's on page 7 of the Differentation pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y ^{n}=u ^{n}v + \left(\begin{matrix}n \\ 1\end{matrix}\right) u ^{n1}v ^{1} + \left(\begin{matrix}n \\ 2\end{matrix}\right) u ^{n2} v ^{2} + ... + uv^{n}\] now that I've worked out the equation editor on here, thats the equation I was referring to.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The part in parentheses is referring to combinations. The whole form is referring to the binomial theorem. Pascal's triangle will also help with this form. Here is a quick link to help with basic examples. http://regentsprep.org/Regents/math/algtrig/ATP4/bintheorem.htm I'll look at your specific problem and get back to you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay...first look for a pattern... take p=1 and q=1 and find the (p+q)derivative or 2nd derivative ...=2 then take p=2 and q= 1 and find the (p+q)derivative or 3rd derivative...=6 note that when n=2 , solution is 2 or 2*1 note that when n=3 , solution is 6 or 3*2*1 therefore make the jump that the solution will be n! (n p)(x^p*(1+x)^q)....=n! Here is another way to think about it.... . For example y=x^2(x+1)^2=x^2(x^2+2x+1)=x^4+2x^3+x^2 Lets take the fourth derivative (the sum of the exponents) y'=4x^3+6x^2+2x y"=12x^2+12x+2 y"'=24x+12 +0 y4=24+0+0=24 note that we multiplied 4*3*2*1=4! in our leading term.... That's all that is happening in the problem....everything else is going to zero... I hope this helps....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, but I'm just wondering what that specific notation of the \[\left(\begin{matrix}n \\ 2\end{matrix}\right)\] or \[\left(\begin{matrix}n \\ p\end{matrix}\right)\]type means in english? Does it just mean we are going to the nth derivative and we are on the second term or the p term or something else?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No its just the combinations...n items taken 2 at a time. Where order does not matter. dw:1327353236094:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n items taken p at a time...sorry. or 4 items taken 2 at a time as presented in the example.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The Binomial Theorem uses this form as well...Most notably used by Pascal and of course the Chinese .... 11 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 ... This represents the coefficients of the binomial powers as listed below (x+1)^1 (x+1)^2 (x+1)^3 (x+1)^4 (x+1)^5 ...
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