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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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perlBest ResponseYou've already chosen the best response.0
maybe you can google a similiar question, and i will reply
 one year ago

perlBest ResponseYou've already chosen the best response.0
ok this is not what i am use to, algebra. what are you studying? what book
 one year ago

perlBest ResponseYou've already chosen the best response.0
engineering mathematics by amit k awasthi
 one year ago

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

perlBest ResponseYou've already chosen the best response.0
what is a factorial polynomial, can you give me a definition
 one year ago

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

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

hartnnBest ResponseYou've already chosen the best response.0
p=2 to make factorial polynomial as quadratic
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
jst a sec. am workng on it
 one year ago

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

hartnnBest ResponseYou've already chosen the best response.0
but that doesn't give u relation between them.......
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
ya i am greatly confused
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
When 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
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
Suppose \(\alpha+\beta x+\gamma x^2=(u+vx)^2\). 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 \(2\sqrt{\alpha\gamma}=\beta\).
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
OH. You're doing Knuthesque math. Okay, one second.
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
The 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 . . .
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
It 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 \(\gamma\).
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
Thank you soooo much for the explanation:)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
is 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
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
I 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?
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
When 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.
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
Does that make things more clear?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
ya it is. thank u soooo much:)
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
You're welcome. But, I'd like to thank you for the interesting problem!
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
Let 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.
 one year ago

LimitlessBest ResponseYou've already chosen the best response.3
You're welcome! Have a wonderful weekend.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
Thanks nd wish u the same:)
 one year ago
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