Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - set f= x y z

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    76

    set f= x y z

    set f = f(x,y,z) = (x^2 - y + 2z^3)^3

    determine the coeffecients wwith which the following terms appear in F by using the multinomial theorem:

    1) x^2 y^2
    2) y^3
    3) x^2 y z^3
    4) y z^6
    5) x y z

    expres the following statements using quantifiers and usual mathematical symbols
    1) the set of integers has no largest element
    2) the differencee of two real numbers is a real number
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mathcore View Post
    set f = f(x,y,z) = (x^2 - y + 2z^3)^3

    determine the coeffecients wwith which the following terms appear in F by using the multinomial theorem:

    1) x^2 y^2
    2) y^3
    3) x^2 y z^3
    4) y z^6
    5) x y z

    expres the following statements using quantifiers and usual mathematical symbols
    1) the set of integers has no largest element
    2) the differencee of two real numbers is a real number

    This looks a lot like homework: what have you done, where are you stuck?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    Posts
    76
    i realised you got to find the coefficients, in the first one i can see x^2 but i cant see any y^2, but maybe it has to do with the ^3 outside of the bracket. the point i am stuck on is the multinomial theorem.

    i have been researching a lot of websites and i dont understand the way they explain it, i can only find general rules which i dont get how to apply.

    i think the answers has something in general to do with expanding the bracker. here is what i tried to prove i am not just being lazy

    i went with: expanding it

    x^6 + y^3 + 2z^9

    and then take out
    (X^4 + y + 2z^9) leaves x^2y^2 so the coefficient is (X^4 + y + 2z^9)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2010
    Posts
    76
    ive even tried tutorvista and it doesnt help me
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mathcore View Post
    set f = f(x,y,z) = (x^2 - y + 2z^3)^3

    determine the coeffecients wwith which the following terms appear in F by using the multinomial theorem:

    1) x^2 y^2
    2) y^3
    3) x^2 y z^3
    4) y z^6
    5) x y z

    expres the following statements using quantifiers and usual mathematical symbols
    1) the set of integers has no largest element
    2) the differencee of two real numbers is a real number


    Well, use the multinomial theorem! (x^2-y+2z^3)^3=\displaystyle{\sum\limits_{k_1+k_2+k_3=3  }\binom{3}{k_1,k_2,k_3}(x^2)^{k_1}(-y)^{k_2}(2z^3)^{k_3}} .

    Soy, what's the coefficient of x^2y^2 ? We'll get this when k_1=1\,,\,k_2=2\Longrightarrow k_3=0 , so the coefficient is

    \binom{3}{1,2,0}=\displaystyle{\frac{3!}{1!2!0!}}=  3 , etc.

    As for the other two questions: I do (1), you do (2):

    \forall x\in\mathbb{Z}\,\,\exists y\in\mathbb{Z}\,\,(y>x)

    Tonio
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2010
    Posts
    76
    youre awesome but what i dont get is how to use the multinomial when there are powers and numbers inside the bracket, i can do (x + y + z)^3 but i dont know what to do with powers inside, also i dont know what it means by terms, i know the nCr from the binomial and i know its about using factorial so that they add up to the n power (which is 3 in my question) but i dont know how to use it

    also i dont get what you mean for i do 1 you do 2 what is the upiside down A ???
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2010
    Posts
    76
    also the y is negative how do u deal with that
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mathcore View Post
    also the y is negative how do u deal with that

    Who cares? If the power of y is odd the negative sign remains, and if the power of y is even it disappears...this is high school stuff!

    Tonio
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mathcore View Post
    youre awesome but what i dont get is how to use the multinomial when there are powers and numbers inside the bracket, i can do (x + y + z)^3 but i dont know what to do with powers inside, also i dont know what it means by terms, i know the nCr from the binomial and i know its about using factorial so that they add up to the n power (which is 3 in my question) but i dont know how to use it

    also i dont get what you mean for i do 1 you do 2 what is the upiside down A ???

    Well, it is now obvious you're out of your depth: how come you try to answer a question asking you to use quantifiers and you ask the above?!?
    Go back to your notes/book, read and then try it again.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Oct 2010
    Posts
    76
    hello

    there is no notes or book

    i want to know, will the coefficients be always 3 or 6 since its 3!/2!1! or 3!/1s or 0s factorial, or will some be negative

    what about the y^3 question, will this one be negative? the coefficient?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Oct 2010
    Posts
    76
    ok heres another thing i dont get:

    number 5, it is asking for the coeff of XYZ, but the powers INSIDE the bracket are already higher than those
    how do you deal with that?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Oct 2010
    Posts
    76
    ") the differencee of two real numbers is a real number"

    ok would you say that xER and yER so X-Y = zER, is that how you do it?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Why not go to this website?

    Type in Expand (x^2-y+2z^3)^3. Click the equals sign at the extreme right of the input box. See what you get.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Oct 2010
    Posts
    76
    the answer being that i hadnt heard of that website

    you're favulous

    what i must ask though: is how come the yz^6 coefficient is -12 and not -6. i made it -6. 3!/2!1! * -1 * 2 = -6, not -12.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    The term containing y^1z^6 is gotten by \frac{3!}{k!\cdot j!}(-y)^k(2z^3)^j so that means k=1~\&~j=2.

    2^2=4~\&~4\cdot 3=12
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Search Tags


/mathhelpforum @mathhelpforum