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Can someone help me with polynomial and rational algebraic functions

  1. May 2, 2004 #1

    JasonRox

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    How can you tell the two apart?

    Here are some examples in the book:

    1. [tex]3x^3 + 2x + 1[/tex]

    2. [tex]3x^2 + (x + 1)^1/2[/tex]

    3. [tex]\frac{2x + 3}{x^2 + 1}[/tex]

    4. [tex](\frac{x}{x + 1})^X[/tex]
     
  2. jcsd
  3. May 3, 2004 #2
    You just need to distribute.

    #2 distributed is:
    [itex] 3x^2 + 1/2x + 1/2 [/itex]

    and #4 distributed is:
    [itex] \frac{x^2}{x+1} [/itex]

    After distributing all four are very obviously different
     
  4. May 3, 2004 #3
    After relooking at your post I realized that you may not be comparing equations but instead classifying them.

    A polynomial can be express as [itex] ax^n + bx^(n-1) + cx^(n-2) + dx^(n-3) … +c [/itex]

    I’d have to look up the definition for a “rational algebraic expression” but going from memory it is any expression that has only algebraic terms?
     
  5. May 3, 2004 #4
    the polynomial 's power must be postive interger . thats the difference
     
  6. May 3, 2004 #5

    uart

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    A Polynomial (in x) is a linear combination of non-negative powers of x.

    A rational algebraic function is just a fraction N(x)/D(x) where N and D are both polynomials.

    In your examples 1. and 2. are polynomials while 3. and 4. are rational algebraic functions.
     
  7. May 4, 2004 #6

    oh? i though the power of polynomial must be interger, i go check
     
  8. May 4, 2004 #7

    uart

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    No need to check you're correct. It was just a slip, I meant to say non-negative integer but only type non-negative. :eek:
     
  9. May 4, 2004 #8

    JasonRox

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    Thanks, guys.
     
  10. May 4, 2004 #9

    krab

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    JonF: use curly brackets to apply something (in this case ^) to an expression.
    [itex] ax^n + bx^{n-1} + cx^{n-2} + dx^{n-3}...+c [/itex]
     
  11. May 4, 2004 #10
    Why would the constant term be equal to the coefficient in front of x^(n - 2)? ;)
     
  12. May 4, 2004 #11
    argh, thank you
     
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