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ajprincess
 4 years ago
Please help:)
Find the relation between \(\alpha\), \(\beta\), \(\gamma\) in the order that \(\alpha+\beta x+\gamma x^2\) may be expressible in one term in the factorial notation.
ajprincess
 4 years ago
Please help:) Find the relation between \(\alpha\), \(\beta\), \(\gamma\) in the order that \(\alpha+\beta x+\gamma x^2\) may be expressible in one term in the factorial notation.

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perl
 4 years ago
Best ResponseYou've already chosen the best response.0maybe you can google a similiar question, and i will reply

perl
 4 years ago
Best ResponseYou've already chosen the best response.0ok this is not what i am use to, algebra. what are you studying? what book

perl
 4 years ago
Best ResponseYou've already chosen the best response.0engineering mathematics by amit k awasthi

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1hmm no. I am nt studying any particular book. this question was given to me by my lecturer.

perl
 4 years ago
Best ResponseYou've already chosen the best response.0what is a factorial polynomial, can you give me a definition

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1A factorial polynomial \(x^p\) is defined as \(x^p=x(xh)(x2h)(x(p1)h)\) where p is a positive integer.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0so, here take p= 2, a+bx+cx^2=x(xh) and find relation between a,b,c.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0p=2 to make factorial polynomial as quadratic

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1jst a sec. am workng on it

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1I am nt sure if what I have done is right. a=0, b=h and c=1. am I right @hartnn?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0but that doesn't give u relation between them.......

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1ya i am greatly confused

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1When I googled my question i found this link. http://acadmedia.wku.edu/Zhuhadar/eBooks/0977858251STATISTICAL.pdf page number 229. i am trying to understand it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Suppose \(\alpha+\beta x+\gamma x^2=(u+vx)^2\). This means \(u^2+2uvx+v^2x^2=\alpha+\beta x+\gamma x^2.\) Equating coefficients, we have that \(u^2=\alpha\), \(2uv=\beta\), and \(v^2=\gamma\). These three equations imply \(2\sqrt{\alpha\gamma}=\beta\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OH. You're doing Knuthesque math. Okay, one second.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The above answer is technically true, but it's not what the professor is looking for. I'll explain what he's doing in a moment . . .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It would appear your professor made a significant error. He's using what's called the rising factorial and incorrectly at that. (He or she, whoever.) The following is the definition of the notation \(a^{(n)}\): \[ a^{(n)}=a(a+1)(a+2)\cdots(a+n1)=\prod_{1\le i\le n}\left(a+i1\right).\] What your professor wants, I believe, is the following: Let \(\alpha+\beta x+\gamma x^2=(u+vx)^{(2)}\). Then, following the definition of the rising factorial, we have \[(u+vx)^{(2)}=\prod_{1 \le i \le 2}\left(u+vx+i1\right)=(u+vx)(u+vx+1).\] Expanding that our, we get \[(u+vx)(u+vx+1)=u^2+uvx+u+uvx+v^2x^2+vx=u^2+u+2uvx+vx+v^2x^2.\] Since we have let \(\alpha+\beta x+\gamma x^2=(u+vx)^{(2)}\), we have that \(\alpha+\beta x+\gamma x^2=(u^2+u)+(2uv+v)x+v^2x^2.\) From this we can conclude (by "equating coefficients") \(\alpha=u^2+u\), \(\beta=2uv+v\), and \(\gamma=v^2\). To search for a relation between our three variables, substitue \(\pm \sqrt{\gamma}\) for \(v\): \(\beta=\pm2u\sqrt{\gamma}\pm\sqrt{\gamma}\). Solving this equation for \(u\), we see that \(u=\frac{\beta \pm \gamma}{\pm 2\sqrt{\gamma}}.\) Substitution of this into \(\alpha=u^2+u\) reveals \[\alpha=\frac{(\beta\pm \gamma)^2}{4\gamma}+\frac{\beta \pm \gamma}{\pm 2\sqrt{\gamma}},\] giving us the desired: a relation between \(\alpha, \beta\) and \(\gamma\).

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1Thank you soooo much for the explanation:)

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1is that way wrong? So which method should I use? Actually the answer I posted is not the answer my professor gave me. Actually she didnt give us any answer yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think the way was entirely wrong. It's just that the author misunderstood the definition of \(a^{(n)}\). You know what I mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When the writer of that post said \[(a+bx)^{(2)}=(a+bx)[a+b(x1)],\] they were wrong and this messed up the entire problem. However, what they were _trying_ to do was right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does that make things more clear?

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1ya it is. thank u soooo much:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You're welcome. But, I'd like to thank you for the interesting problem!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let me go ahead and recommendif you're getting problems like this a lotDonald E. Knuth's Concrete Mathematics. In there, there's all kinds of this craziness: falling factorials, rising factorials, ceiling functions, floor functions, summations, discrete calculus, etc. It seems to directly pertain to what you're doing, but I can't be certain.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You're welcome! Have a wonderful weekend.

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1Thanks nd wish u the same:)
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