Question

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!

Answer 1

Do you have a link to the problem set or can you post the problem?

Answer 2

I assume, this is the link...
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/assignments/

Answer 3

I didn't find the problem...please post.

Answer 4

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

Answer 5

\[y ^{n}=u ^{n}v + \left(\begin{matrix}n \\ 1\end{matrix}\right) u ^{n-1}v ^{1} + \left(\begin{matrix}n \\ 2\end{matrix}\right) u ^{n-2} v ^{2} + ... + uv^{n}\]

now that I've worked out the equation editor on here, thats the equation I was referring to.

Answer 6

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.

Answer 7

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....

Answer 8

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?

Answer 9

No its just the combinations...n items taken 2 at a time. Where order does not matter.