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Coefficient Of Expansion

  1. Jul 31, 2008 #1

    Air

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    1. The problem statement, all variables and given/known data
    a) Find the coefficient of [itex]x^5[/itex] in [itex](1+x+x^2)^4[/itex].

    b) Find the coefficient of [itex]x^2[/itex] in [itex](2+2x+x^2)^n[/itex].


    2. The attempt at a solution
    a) [itex] (1)^0(x)^3(x^2)^1[/itex] & [itex](1)^1(x)^1(x^2)^2[/itex].
    Coefficient of first: [itex]\frac{4!}{(0!)(3!)(1!)} = 4[/itex]
    Coefficient of second: [itex]\frac{4!}{(1!)(1!)(2!)}=16[/itex]
    Therefore coefficient of [itex]x^5[/itex] is [itex]4+12 = 16[/itex].

    b) This is the part I'm struggling on. May I have help here. It's a general term (power of [itex]n[/itex] which is confusing me).
     
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  3. Jul 31, 2008 #2

    tiny-tim

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    Hi Air! :smile:

    (btw, "Coefficient Of Expansion" means something specific in materials physics)

    Well, it's either x2 or x x;

    if you use the same formulas as before, what do you get? :smile:
     
  4. Jul 31, 2008 #3

    Air

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    So would that be:

    [itex](2)^0(2x)^0(x^2)^1[/itex] and [itex](2)^1(2x)^2(x^2)^0[/itex]

    What am I to do with the [itex](2)[/itex]? It can be used many times? :confused:

    ("Coefficient Of Expansion" was just a random title I thought of which might be suitable for this, had no idea that it's something in materials physics :smile:)
     
  5. Jul 31, 2008 #4

    tiny-tim

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    No, it's …

    [tex](2)^{n-1}(2x)^0(x^2)^1[/tex] and … ? :smile:
     
  6. Jul 31, 2008 #5

    Air

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    Hmmm.... I'm not sure how it's possible to get [itex](2x)^2[/itex] and [itex](x^2)^0[/itex] together as the powers have to be consecutive, don't they?
     
  7. Jul 31, 2008 #6

    tiny-tim

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    erm … I've no idea what you mean … but the answer is, no! :smile:
     
  8. Jul 31, 2008 #7

    Air

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    Would it be this:

    [itex](2)^{n}(2x)^2(x^2)^0[/itex]

    I'm sure about the [itex](2x)^2(x^2)^0[/itex] because it would get [itex](4x^2)(1)[/itex] which would give a coefficient of value [itex]4[/itex] but is the power of [itex]2[/itex] correct?
     
  9. Jul 31, 2008 #8

    tiny-tim

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    Hint: the exponents (that's the itsy-bitsy thingys) have to add to n, don't they? :smile:

    (Yours are adding to n+2.)
     
  10. Jul 31, 2008 #9

    HallsofIvy

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    Don't be sure- that term won't be in an nth power. In [itex](2+2x+x^2)^n[/itex], every term is of the form Cijk(2)i(2x)j(x2)k with i+ j+ k= n and C is the "tri"-nomial coefficient:
    [tex]C_{ijk}= \frac{n!}{i!j!k!}[/tex]

    One way to get x2 would be (2)n-2(2x)2(x2)0 but another would be (2n-1)(2x)0(x2)1. And don't forget the "tri"-nomial coefficient for each of those.
     
    Last edited: Jul 31, 2008
  11. Jul 31, 2008 #10

    Air

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    Oh, I see. Thanks. That makes sense.

    But I have another question:
    As this question is [itex](2+2x+x^2)^n[/itex], does that mean it's even possible to find the coefficient of [itex]x^{235}[/itex] (for instance) as [itex]n[/itex] can be any value?
     
  12. Jul 31, 2008 #11

    HallsofIvy

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    Yes, it certainly should be. As I said before, Cijk(2)i(2x)j(x[sup2)k with i+ j+ k= n and C is the "tri"-nomial coefficient.

    In order to get x235, you would have to have j+ 2k= 235. There are many values of n large enough to include that. Of course, with x2 any positive n is sufficient.
     
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