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ajprincess

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

    • one year ago
  2. perl
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    ok this is not what i am use to, algebra. what are you studying? what book

    • one year ago
  3. perl
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    engineering mathematics by amit k awasthi

    • one year ago
  4. ajprincess
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    hmm no. I am nt studying any particular book. this question was given to me by my lecturer.

    • one year ago
  5. perl
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    what is a factorial polynomial, can you give me a definition

    • one year ago
  6. ajprincess
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    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.

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

    • one year ago
  8. hartnn
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    but i am not sure...

    • one year ago
  9. hartnn
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    p=2 to make factorial polynomial as quadratic

    • one year ago
  10. ajprincess
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    jst a sec. am workng on it

    • one year ago
  11. ajprincess
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    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
  12. hartnn
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    but that doesn't give u relation between them.......

    • one year ago
  13. ajprincess
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    ya i am greatly confused

    • one year ago
  14. ajprincess
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    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

    • one year ago
  15. ajprincess
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    • one year ago
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  16. Limitless
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    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
  17. Limitless
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    OH. You're doing Knuth-esque math. Okay, one second.

    • one year ago
  18. Limitless
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    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
  19. Limitless
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    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 \(\gamma\).

    • one year ago
  20. ajprincess
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    Thank you soooo much for the explanation:)

    • one year ago
  21. ajprincess
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    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
  22. Limitless
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    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
  23. ajprincess
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    hmm no. sorry

    • one year ago
  24. Limitless
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    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.

    • one year ago
  25. Limitless
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    Does that make things more clear?

    • one year ago
  26. ajprincess
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    ya it is. thank u soooo much:)

    • one year ago
  27. Limitless
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    You're welcome. But, I'd like to thank you for the interesting problem!

    • one year ago
  28. Limitless
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    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.

    • one year ago
  29. ajprincess
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    Thanks a lott:)

    • one year ago
  30. Limitless
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    You're welcome! Have a wonderful weekend.

    • one year ago
  31. ajprincess
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    Thanks nd wish u the same:)

    • one year ago
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