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Throol
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Could someone please explain problem 1G5b of the first problem set? I looked online for the binomial theorem and how to solve binomial coefficients, and I've got the first part done (mostly guessing), totally lost on the part with y^(p+q).
 3 years ago
 3 years ago
Throol Group Title
Could someone please explain problem 1G5b of the first problem set? I looked online for the binomial theorem and how to solve binomial coefficients, and I've got the first part done (mostly guessing), totally lost on the part with y^(p+q).
 3 years ago
 3 years ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
without having any of the materials you mention, i gots no way of knowing what it is you actually need ...
 3 years ago

FabianMontescu Group TitleBest ResponseYou've already chosen the best response.1
This is mighty tricky, and the answer sheet (attached and also available on the OCW site) has a typo  k inside the binomial (n k) should be p. If you apply Leibniz formula, you'll find that the first term is: \[\left(\begin{matrix}n \\ 0\end{matrix}\right) (x^p)^{(p +q)}(1+x)^q \] The thing to notice is that if you derive x^p more than p times, the derivative is going to be zero. Since you're deriving it p+q times, the derivative is going to be zero. A similar argument can be applied to all terms except one, the term that is: \[\left(\begin{matrix}n \\ p\end{matrix}\right) (x^p)^{(p)}[(1+x)^q]^{(q)} \] Therefore, all the terms except the one above are zero. Taking the derivative of that term is not really pretty, but from the end of lecture 4 (or from trying it yourself) you'll find that the derivative of \[(x^p)^{(p)} = p! \] Similarly, using the chain rule (try it with q = 2 or q =3 so you can see it) \[[(1+x)^q]^{(q)} = q! \] So the nonzero term is \[\left(\begin{matrix}n \\ p\end{matrix}\right) p!q! \] There's one more step: you have to note that n = p+q, so you can solve the binomial. You'll get (for the binomial) \[ \left(\begin{matrix}n \\ p\end{matrix}\right) = \frac{(p+q)!}{p!(p+qp)!} = \frac{(p+q)!}{p!q!} = \frac{n!}{p!q!}\] Substituting this in the nonzero term will get you: \[ \frac{n!}{p!q!}p!q! = n!\] This was kinda hard, but I hope it helped.
 2 years ago

Throol Group TitleBest ResponseYou've already chosen the best response.1
How do you get from the equation given in the problem set, \[y ^{n}=u ^{(n)}v+\left(\begin{matrix}n \\1 \end{matrix}\right)u ^{(n1)}v ^{(1)}+\left(\begin{matrix}n \\2 \end{matrix}\right)u ^{(n2)}v ^{(2)}+...+uv ^{(n)}\] to the assumption that all terms but one go to zero? I understand that deriving \[x ^{p}\] more than p times results in zero, but why would that be the case for the other terms? i.e., if I substitute n=p+q, \[u=x ^{p}\], and \[v=(1+x)^{q}\] into the equation above, I get \[y ^{(p+q)}=(x ^{p})^{(p+q)}(1+x)^{q}+\left(\begin{matrix}n \\ 1\end{matrix}\right)(x ^{p})^{(p+q1)}((1+x)^{q})^{1}+\left(\begin{matrix}n \\ 2\end{matrix}\right)(x ^{p})^{(p+q2)}((1+x)^{q})^{(2)}\] \[+...+x ^{p}((1+x)^{q})^{(p+q)}\] The first term should be zero since p+q>p (assuming q>0), but how do you assume that p+q1>p or that p+q2>p, etc.? Do we assume that q doesn't equal 1 or 2, etc.? How does one go from the given information to deciding to find the (n p) term specifically?
 2 years ago

Throol Group TitleBest ResponseYou've already chosen the best response.1
Oops, I forgot to substitute the n's in the (n 1) etc. terms, should instead be ((p+q) 1), etc.
 2 years ago

FabianMontescu Group TitleBest ResponseYou've already chosen the best response.1
Let's start with an example. If p = 4 and q = 2, p + q > p, p+q1 > p, p + q  2 = p and p + q  3 < p. You can easily see that the first two are going to become zero, the third (based on the discussion above) is going to remain and the last one is the interesting case. Although the derivative of p exists on that case, the derivative of (1+x)^q will be zero, as q=2 but we're taking the third derivative. More generally, for 0 < i < p + q = n we have that: \[ p + q  i < p \Rightarrow \ i > q \]\[ p + q  i = p \Rightarrow \ i = q \] So we have three cases: 1) The derivative of x^p has an order higher than p. The whole term is zero. 2) The derivative of x^p has a lower order than p. Then the derivative of (1+x)^q has an order that is higher than q. The whole term is zero. 3) We're taking the pth derivative of x^p, and therefore we're taking the qth derivative of (1+x)^q. This term remains, and it has the formula of my first answer. Hope this helps, but I can try again if it didn't!
 2 years ago

Throol Group TitleBest ResponseYou've already chosen the best response.1
Ooh I hate this software. I had a response all prepared, then the login popped up, I closed it, and it erased everything...
 2 years ago

Throol Group TitleBest ResponseYou've already chosen the best response.1
Ok, here I go again: For the ith term, \[\left(\begin{matrix}(p+q) \\ i\end{matrix}\right)(x ^{p})^{(p+qi)}((1+x)^{q})^{(i)}\] As you explained, 0<i<(p+q)=n, p+qi<p=>i>q, so the (1+x)^q term would go to zero p+qi=p=>i=q Also to finish it, p+qi>p, so the x^p term would go to zero. Makes sense. So the only term that doesn't cancel is i=q, \[\left(\begin{matrix}(p+q) \\ q\end{matrix}\right)(x ^{p})^{(p)}((1+x)^{q})^{(q)}\] This makes sense, and fits your statement 3. But this matches neither the answer sheet's answer (which must be a typo), nor your original equation of \[\left(\begin{matrix}n \\ p\end{matrix}\right)(x ^{p})^{(p)}((1+x)^{q})^{(q)}\] If I'm following this correctly, it looks like we want the qth term, not the pth term, right?
 2 years ago

FabianMontescu Group TitleBest ResponseYou've already chosen the best response.1
Yes, I should have used the q term, it would have made it easier to understand. The answer key is correct, even though it's not intuitive: \[\text{When } p+q=n \text{, then } \left(\begin{matrix}n \\ p\end{matrix}\right) = \left(\begin{matrix}n \\ q\end{matrix}\right)\] Sorry about that, it would have been easier to have used q from the beginning.
 2 years ago

Throol Group TitleBest ResponseYou've already chosen the best response.1
Ok, I calculated it all out, and it works! Thanks!!!
 2 years ago
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