## ajprincess Group Title 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. 2 years ago 2 years ago

1. perl

2. perl

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

3. perl

engineering mathematics by amit k awasthi

4. ajprincess

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

5. perl

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

6. perl
7. 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.

8. hartnn

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

9. hartnn

but i am not sure...

10. hartnn

p=2 to make factorial polynomial as quadratic

11. ajprincess

jst a sec. am workng on it

12. ajprincess

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

13. hartnn

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

14. ajprincess

ya i am greatly confused

15. ajprincess

16. ajprincess

17. Limitless

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 $$2\sqrt{\alpha\gamma}=\beta$$.

18. Limitless

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

19. Limitless

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 . . .

20. Limitless

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 $$\gamma$$.

21. ajprincess

Thank you soooo much for the explanation:)

22. 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

23. Limitless

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?

24. ajprincess

hmm no. sorry

25. Limitless

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.

26. Limitless

Does that make things more clear?

27. ajprincess

ya it is. thank u soooo much:)

28. Limitless

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

29. Limitless

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.

30. ajprincess

Thanks a lott:)

31. Limitless

You're welcome! Have a wonderful weekend.

32. ajprincess

Thanks nd wish u the same:)