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

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- katieb

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- perl

maybe you can google a similiar question, and i will reply

- perl

ok this is not what i am use to, algebra. what are you studying? what book

- perl

engineering mathematics by amit k awasthi

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## More answers

- ajprincess

hmm no. I am nt studying any particular book. this question was given to me by my lecturer.

- perl

what is a factorial polynomial, can you give me a definition

- perl

http://books.google.com/books?id=bwZKJ95CVTwC&pg=PA314&lpg=PA314&dq=Find+the+relation+between+%CE%B1,+%CE%B2,+%CE%B3+in+the+order+that+%CE%B1%2B%CE%B2x%2B%CE%B3x2+may+be+expressible+in+one+term+in+the+factorial+notation.&source=bl&ots=HBfAzq833r&sig=dcBssAwJZJpHrv1BK7f1RAZH1no&hl=en&sa=X&ei=eIOwUNHfEcSx0AH-nIDIBw&ved=0CC8Q6AEwAA#v=onepage&q=Find%20the%20relation%20between%20%CE%B1%2C%20%CE%B2%2C%20%CE%B3%20in%20the%20order%20that%20%CE%B1%2B%CE%B2x%2B%CE%B3x2%20may%20be%20expressible%20in%20one%20term%20in%20the%20factorial%20notation.&f=false

- ajprincess

A factorial polynomial \(x^p\) is defined as
\(x^p=x(x-h)(x-2h)--------------(x-(p-1)h)\)
where p is a positive integer.

- hartnn

so, here take p= 2,
a+bx+cx^2=x(x-h)
and find relation between a,b,c.

- hartnn

but i am not sure...

- hartnn

p=2 to make factorial polynomial as quadratic

- ajprincess

jst a sec. am workng on it

- ajprincess

I am nt sure if what I have done is right.
a=0, b=-h and c=1. am I right @hartnn?

- hartnn

but that doesn't give u relation between them.......

- ajprincess

ya i am greatly confused

- ajprincess

When I googled my question i found this link.
http://acadmedia.wku.edu/Zhuhadar/eBooks/0977858251-STATISTICAL.pdf
page number 229. i am trying to understand it

- ajprincess

##### 1 Attachment

- anonymous

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 \(2\sqrt{\alpha\gamma}=\beta\).

- anonymous

OH. You're doing Knuth-esque math. Okay, one second.

- anonymous

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 . . .

- anonymous

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 \(a^{(n)}\):
\[ a^{(n)}=a(a+1)(a+2)\cdots(a+n-1)=\prod_{1\le i\le n}\left(a+i-1\right).\]
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 \[(u+vx)^{(2)}=\prod_{1 \le i \le 2}\left(u+vx+i-1\right)=(u+vx)(u+vx+1).\]
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

Thank you soooo much for the explanation:)

- ajprincess

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

- anonymous

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?

- ajprincess

hmm no. sorry

- anonymous

When the writer of that post said
\[(a+bx)^{(2)}=(a+bx)[a+b(x-1)],\]
they were wrong and this messed up the entire problem. However, what they were _trying_ to do was right.

- anonymous

Does that make things more clear?

- ajprincess

ya it is. thank u soooo much:)

- anonymous

You're welcome. But, I'd like to thank you for the interesting problem!

- anonymous

Let me go ahead and recommend--if you're getting problems like this a lot--Donald 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.

- ajprincess

Thanks a lott:)

- anonymous

You're welcome! Have a wonderful weekend.

- ajprincess

Thanks nd wish u the same:)

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