anonymous
  • anonymous
Can anyone help me with a notation question? In Problem Set 1, 1G-5 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!
MIT 18.01 Single Variable Calculus (OCW)
chestercat
  • chestercat
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anonymous
  • anonymous
Do you have a link to the problem set or can you post the problem?
anonymous
  • anonymous
I assume, this is the link... http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/assignments/
anonymous
  • anonymous
I didn't find the problem...please post.

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anonymous
  • anonymous
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/problem-set-1/ It's on page 7 of the Differentation pdf
anonymous
  • anonymous
\[y ^{n}=u ^{n}v + \left(\begin{matrix}n \\ 1\end{matrix}\right) u ^{n-1}v ^{1} + \left(\begin{matrix}n \\ 2\end{matrix}\right) u ^{n-2} v ^{2} + ... + uv^{n}\] now that I've worked out the equation editor on here, thats the equation I was referring to.
anonymous
  • anonymous
The 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
  • anonymous
okay...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
  • anonymous
Okay, 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 n-th derivative and we are on the second term or the p term or something else?
anonymous
  • anonymous
No its just the combinations...n items taken 2 at a time. Where order does not matter. |dw:1327353236094:dw|
anonymous
  • anonymous
n items taken p at a time...sorry. or 4 items taken 2 at a time as presented in the example.
anonymous
  • anonymous
The 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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