Results 1 to 4 of 4

Math Help - Problem 45

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9

    Problem 45

    1)Prove that \sin 1^{\circ} and \cos 1^{\circ} are irrational.

    3)Suppose that \sum_{k=0}^n k^m = P_m(n) where P_m is an m+1-degree polynomial. Prove that the leading coefficient of P_m is 1/(m+1).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    651
    Thanks
    1
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    1)Prove that \sin 1^{\circ} and \cos 1^{\circ} are irrational.
    Suppose we let:

    f = cos(x),g = sin(x)

    Then:

    e^{xi} = cos(x) + i*sin(x) = f + ig

    Therefore:

     e^{5xi} = cos(5x) + i*sin(5x) = (f + ig)^5

    (f + ig)^5 = f^5 + 5f^4gi - 10f^3g^2 - 10f^2g^3i + 5fg^4 + g^5

    Now we take the real parts:

    cos(5x) = f^5 - 10f^3g^2 + 5fg^4

    Using g^2 = 1 - f^2 throughout:

    cos(5x) = 16f^5 - 20f^3 + 5f

    We know that:

    cos(45-30) = cos(45)cos(30) + sin(45)sin(30) = \frac{\sqrt{3} + 1}{2\sqrt{2}}

    This is clearly irrational, therefore:

    cos(15) = 16cos(3) - 20cos(3) + 5cos(3) = \frac{\sqrt{3} + 1}{2\sqrt{2}}

    If cos(3) was rational, 16cos(3) - 20cos(3) + 5cos(3) would also be rational, which is a contradiction.

    Finally:

    Using cos(3x) = 4cos^3\!(x) - 3cos(x)

    We get:

    cos(3) = 4cos^3\!(1) - 3cos(1)

    If cos(3) was rational, then 4cos^3\!(1) - 3cos(1) would also be rational, and if cos(1) was rational, then the statement above would also be rational, we have thus arrived at a contradiction:

    cos(1) is irrational.

    Q.E.D.

    I have to go to class someone else do sin(1) or I'll do it later.
    Last edited by Aryth; February 13th 2008 at 10:29 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    651
    Thanks
    1
    Awards
    1
    For sin(1) = irrational

    Using the math above and taking the imaginary parts:

    isin(5x) = 5f^4gi - 10f^2g^3i

    Using f^2 = 1 - g^2

    sin(5x) = 5g^9 - 10g^7 + 15g^5 - 5g

    We know that:

    sin(45 - 30) = sin(45)cos(30) + cos(45)sin(30) = \frac{\sqrt{3} + 1}{2}

    That is clearly irrational.

    sin(15) = 5sin^9\!(3) - 10sin^7\!(3) + 15sin^5\!(3) - 5sin(3) = \frac{\sqrt{3} + 1}{2}

    sin(3) is irrational since sin(15) is irrational.

    Finally:

    sin(3x) = 3sin(x) - 4sin^3\!(x)

    Therefore:

    sin(3) = 3sin(1) - 4sin^3\!(1)

    Therefore, since sin(3) is irrational, sin(1) is also irrational.

    Q.E.D.

    Using this as logic:

    sin(nx) \ is \ irrational  \Rightarrow sin(x) \ is \ also \ irrational

    Same with cos(nx)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    2)We are given that \sum_{k=1}^n k^m = P_m(n) where P_m is a polynomial. Suppose we wish to compute the integral of f(x) = x^m over the interval [0,1]. Then \int_0^1 x^m ~ dx = \frac{1}{m+1}. Now there is another way to compute this integral, and that is to use a Riemann sum with n equal subdivision point using the right endpoint. In that case we get \lim ~ \left( \sum_{k=1}^n \frac{k^m}{n^m} \right) \cdot \frac{1}{n} = \lim ~ \frac{1}{n^{m+1}} \sum_{k=1}^n k^m = \lim ~ \frac{P_m(n)}{n^{m+1}}. This limit needs to be 1/(m+1) since that is the integral. But this means the polynomial P_m(n) must be degree m+1 with leading coefficient of 1/(m+1).


    In fact, note the following,
    \sum_{k=1}^n k = \frac{n(n+1)}{2} = \boxed{\frac{1}{2}}n^2 + \frac{1}{2}n.
    \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} = \boxed{\frac{1}{3}}n^3 + \frac{1}{2}n^2+\frac{1}{6}n.
    \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4} = \boxed{\frac{1}{4}}n^4 + \frac{1}{2}n^3+\frac{1}{4}n^2.
    ......
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum