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  1. Top | #21
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    Quote Originally Posted by Bomb#20 View Post
    Quote Originally Posted by beero1000 View Post
    1. If x,y,z are positive real numbers with xyz = 1, show that x2+y2+z2 ≤ x3+y3+z3
    Assume without loss of generality that x <= y <= z. z = 1/xy. If all three were less than 1 their product would be less than 1; likewise, if all three were greater than 1 their product would be greater than 1. Therefore x <= 1 and z >= 1. There are two cases to consider, y <= 1 and y > 1. Here's a proof for y <= 1.



    To show that x2+y2+z2 <= x3+y3+z3 is to show that f = x3+y3+z3 - (x2+y2+z2) >= 0.

    Let r = 1 - x, s = 1 - y, t = z - 1. r, s, t >= 0.

    x(1 + r/x) = x + r = x + 1 - x = 1. Therefore 1 + r/x = 1/x. Similarly, 1 + s/y = 1/y.

    z = 1/xy = (1/x)(1/y) = (1 + r/x)(1 + s/y) = 1 + r/x + s/y + rs/xy

    t = r/x + s/y + rs/xy

    f = x3+y3+z3 - (x2+y2+z2)

    f = (1-r)3+(1-s)3+(1+t)3 - ((1-r)2+(1-s)2+(1+t)2)

    f = (3 + 3(t-r-s) + 3(t2+r2+s2) + t3-r3-s3)
    - (3 + 2(t-r-s) + t2+r2+s2)

    f = t-r-s + 2(t2+r2+s2) + t3-r3-s3

    Let g = 2(t2+r2+s2)

    f = r/x + s/y + rs/xy - r - s + g + (r/x + s/y + rs/xy)3-r3-s3

    (r/x + s/y + rs/xy)3 = (r/x)3 + (s/y)3 + [a series of 25 more products of sets of three terms].

    Let h = that series of 25 products I'm not going to painfully write out.

    f = r/x + s/y + rs/xy - r - s + g + (r/x)3 + (s/y)3 + h - r3 - s3

    f = (r/x - r) + (s/y - s) + rs/xy + g + ((r/x)3 - r3) + ((s/y)3 - s3) + h

    f = r(1/x - 1) + s(1/y - 1) + rs/xy + g + r3((1/x)3 - 1) + s3((1/y)3 - 1) + h

    f = r(1 + r/x - 1) + s(1 + s/y - 1) + rs/xy + g + r3((1 + r/x)3 - 1) + s3((1 + s/y)3 - 1) + h

    f = r(r/x) + s(s/y) + rs/xy + g + r3((1 + r/x)3 - 1) + s3((1 + s/y)3 - 1) + h

    By inspection, every individual term in the above sum of terms is non-negative. Therefore f >= 0. Q.E.D.



    The proof for y > 1 is analogous and is left as an exercise for the masochist.
    Ouch!

  2. Top | #22
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    Quote Originally Posted by beero1000 View Post
    The "sneaky" approach:

    Note that the order of the sides does not matter, so take the octagon side sequence to be alternating 4's and 2's. Take a square of side s and symmetrically chop off the corners at 45 degrees. You are left with an octagon that has two groups of four equal edges, and it is clearly cyclic. We want the small sides to have length 2, so we cut \sqrt{2} off each edge. Then we want s - 2 \sqrt{2} = 4, so s = 4 + 2\sqrt{2}. Therefore, the area is (4+2\sqrt{2})^2 - 4 = 20 + 16\sqrt{2}.

    The circle's radius is then \sqrt{(2+\sqrt{2})^2 + 2^2} = \sqrt{10 + 4\sqrt{2}}
    That's a lot smarter than what I did.

    Quote Originally Posted by beero1000 View Post
    Quote Originally Posted by Bomb#20 View Post
    ...
    The proof for y > 1 is analogous and is left as an exercise for the masochist.
    Ouch!
    Ouch indeed. So is it safe to assume there's a "sneaky" approach to the xyz=1 problem too?

  3. Top | #23
    Quantum Hot Dog Kharakov's Avatar
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    Quote Originally Posted by Bomb#20 View Post
    Quote Originally Posted by beero1000 View Post
    1. If x,y,z are positive real numbers with xyz = 1, show that x2+y2+z2 ≤ x3+y3+z3
    Assume without loss of generality that x <= y <= z. z = 1/xy. If all three were less than 1 their product would be less than 1; likewise, if all three were greater than 1 their product would be greater than 1. Therefore x <= 1 and z >= 1. There are two cases to consider, y <= 1 and y > 1. Here's a proof for y <= 1.

    The proof for y > 1 is analogous and is left as an exercise for the masochist.
    That helped.


    I do it the opposite way x>=y>=z, which may be a character flaw... anyway.

    x^2 (x-1) + y^2 (y-1) + z^2 (z-1) \geq 0

    if x and y are greater than 1, the only negative is the z term, and it will be smaller than 1. whoops...

    If x>1 and y = 1, we only have the x and z terms to worry about. The x term will be larger because squaring increases the value of the x term, and decreases the value of the z term.

    if only x is greater than 1, the y and z portions will both be negative. y=1/xz z= 1/xy
    Working on it.

    Last edited by Kharakov; 04-20-2015 at 05:44 AM.

  4. Top | #24
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    Quote Originally Posted by Bomb#20 View Post
    That's a lot smarter than what I did.

    Quote Originally Posted by beero1000 View Post
    Quote Originally Posted by Bomb#20 View Post
    ...
    The proof for y > 1 is analogous and is left as an exercise for the masochist.
    Ouch!
    Ouch indeed. So is it safe to assume there's a "sneaky" approach to the xyz=1 problem too?
    Safe enough.


    Based on two simple and very useful inequalities:
    1. AM-GM: \sqrt[3]{xyz} \leq (x+y+z)/3
    2. Rearrangement inequality: A sum of products is maximized when the factors are matched in order.


    By the AM-GM inequality:
    x^2 + y^2 + z^2 = (x^2 + y^2 + z^2)(\sqrt[3]{xyz}) \leq (x^2 + y^2 + z^2)(x+y+z)/3

    Multiplying through and using the rearrangement inequality on the factors x^2,y^2,z^2 and x,y,z:
    (x^3 + y^3 + z^3) + (x^2y + y^2z + z^2x) + (x^2z + y^2x + z^2y) \leq 3(x^3 + y^3 + z^3)

    Put them together to get x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3.

    In general, this technique shows that if x_1x_2 \dots x_k = 1 then x_1^n + x_2^n + \dots + x_k^n \leq x_1^m + x_2^m + \dots + x_k^m for m \geq n > 0.


  5. Top | #25
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    Quote Originally Posted by beero1000 View Post
    3. What is \displaystyle \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} \frac{\sin x}{x^3} dx?
    Taking the Taylor expansion of sin(x) around x = 0,

    \displaystyle \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} \frac{\sin x}{x^3} dx = \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!} \frac{x^{2m+1}}{x^3} dx
    = \lim_{n\to\infty}\sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!}  \int_{1/(n+1)}^{1/n} x^{2m - 2}\, dx

    Since the integration limits both go to zero, the only contribution can be from the terms that approach a singularity as x goes to zero. Therefore, only the m = 0 term contributes, giving
    \displaystyle = \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} x^{- 2}\, dx =\lim_{n\to\infty} (-x)^{-1} |_{x = 1/(n+1)}^{1/n} = \lim_{n\to\infty}-[n - (n+1)] = 1.
    Last edited by Artemus; 04-20-2015 at 05:08 PM.

  6. Top | #26
    Administrator lpetrich's Avatar
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    I came rather later to this party, so I've ended up duplicating others' solutions.

    Quote Originally Posted by beero1000 View Post
    Suppose the roots of the polynomial x5 + 2x4 + 3x3 + 4x2 + 5x + 6 are a, b, c, d, and e, respectively. Find a2+b2+c2+d2+e2. Bonus points: Find the harmonic mean of the roots (i.e. 5 times the reciprocal of the sum of their reciprocals).
    One can avoid solving the equation by using Newton's identities.

    Notation:
    en = sub of n x's all distinct -- these appear in the polynomial's coefficients
    pn = sum of the nth power of each x

    We want p2:
    e1 = - 2
    e2 = 3
    p1 = e1
    p2 = e1*p1 - 2*e2
    Thus,
    p1 = -2
    p2 = -2

    For the harmonic mean, we find the polynomial for 1/x to get the reciprocals of the original one's roots:

    (1/x)5 + (5/6)(1/x)4 + (2/3)(1/x)3 + (1/2)(1/x)2 + (1/3)(1/x) + (1/6)
    e1 = -5/6
    Mean = 1/5 of this, because it has 5 roots, or -1/6
    Reciprocal of that mean = harmonic mean of original roots = -6

    An irregular octagon is inscribed in a circle. The edge lengths of the octagon are 4,4,4,4,2,2,2,2. Find the area of the octagon. Bonus points: Find the radius of the circle.
    The solution turns out to be independent of their order in the circle. For side length s and radius r, the angle subtended by it at the origin is 2*arcsin(s/(2r)). So, arcsin(1/r) + arcsin(2/r) = pi/4.

    The area of each side triangle: (s/2)*sqrt(r^2 - (s/2)^2)

    Radius = sqrt(10 + 4*sqrt(2))

    Area = 4 * (2*sqrt(2) + 1) + 4 * (2 * (2 + sqrt(2))) = 20 + 16*sqrt(2) = 4 * (5 + 4*sqrt(2))

    Is (14n+3)/(21n + 4) a fraction in simplest form for every n?
    Presumably integer n.

    Let's use Euclid's algorithm for the GCD:
    21n + 4
    14n + 3
    7n + 1
    1

    The numerator and the denominator are thus always relatively prime, making the fraction always in lowest terms.

    Two unit circles pass through each other's center. Find the area of their intersection. Bonus points: Same question for 3 intersecting circles.
    If their centroid is at {0,0} and their centers are at {1/2,0}, and {-1/2,0}, then their intersection points are at {0,sqrt(3)/2} and {0,-sqrt(3)/2}

    The circle slice from center to each intersection point has angle 2*60d = 120d = 2pi/3. That gives an area of (1/2)*this or pi/3 The triangle between those points has area sqrt(3)/4. Their difference: pi/3 - sqrt(3)/4. The intersection region is twice this, with area
    2*pi/3 - sqrt(3)/2 ~ 1.22837


    For the 3-circle case, we take one of the circles, with center {-1/2, 0}. The part of it shared by the others is a circle slice with angle pi/3 - triangle {-1/2,0} - {1/2,0} - {0,sqrt(3)/2} + triangle {1/2,0} - {0,sqrt[3}/2} - {0,sqrt{3}/6}

    That's pi/6 - sqrt(3)/4 + sqrt(3)/12 = pi/6 - sqrt(3)/6

    For all three together, we get
    pi/2 - sqrt(3)/2 ~ 0.7046

    Both results were checked numerically.

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    Administrator lpetrich's Avatar
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    Quote Originally Posted by beero1000 View Post
    f x,y,z are positive real numbers with xyz = 1, show that x2+y2+z2 ≤ x3+y3+z3
    One can solve this by finding the minimum of
     (x^3+y^3+z^3) - (x^2+y^2+z^2) - \lambda xyz

    putting in the constraint with a Lagrange multiplier. Taking derivatives gives
     x^2(3x-2) = y^2(3y-2) = z^2(3z-2) = \lambda

    Crunching with Mathematica yields only one real solution: x = y = z = 1.

    Since the derivatives are nonzero elsewhere, the inequality is always in one direction elsewhere. For x = 2, y = 1/2, z = 1, the lhs is 21/4, and the rhs is 73/8, and rhs > lhs. Thus, rhs > lhs everywhere.

    If n is a positive even number, is 23n-1 + 5(3n) divisible by 11?
    Consider the function f(n) = 26n-1 + 5*32n

    It satisfies recurrence relation f(n+2) -73*f(n+1) + 576*f(n) = 0 -- it's a 3-term one because the function has 2 exponentials in it.

    So if any two consecutive ones are evenly divisible by 11, all the rest are. f(1) = 77 = 7*11, f(2) = 2453 = 223*11, and thus all the rest are.

    What is \displaystyle \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} \frac{\sin x}{x^3} dx?
    Since x is small, it becomes \displaystyle \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} \frac{1}{x^2} + O(1) \, dx = \lim_{n\to\infty} (n+1) - n + O(1/n^2) = 1

    Find the surface area and volume of a horn torus (a circle rotated about one of its tangents).
     \displaystyle A = 2 \pi \int \sqrt{ 1 + \left( \frac{dr}{dz} \right)^2} r \, dz
     \displaystyle V = \pi \int (r_{outer}(z)^2 - r_{inner}(z)^2) \, dz

    For a torus with circle-center-path radius a and circle radius b,

     A = 4\pi^2 a b ,\ V = 2\pi a b^2

    Here, a = b.

  8. Top | #28
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    Bored today and thinking about spirals. Find the arc-length formulas for Archimedes' spiral (r = c θ):
    1. On the Euclidean plane
    2. On the Elliptic plane
    3. On the Hyperbolic plane

    Solver's choice on details for the metrics/models and FYI, calculation-wise, these are not as nice as my usual questions...

  9. Top | #29
    Quantum Hot Dog Kharakov's Avatar
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    I solved the |c|=0 case. I'll leave the |c| > 0 case for the rest of you.



    1.  c (\theta \sin(\theta)+\sin(\theta)-\theta \cos(\theta)+\cos(\theta))


    Last edited by Kharakov; 04-26-2015 at 06:31 AM.

  10. Top | #30
    Administrator lpetrich's Avatar
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    First, LaTeX/Mathematics - Wikibooks, open books for an open world goes into detail about LaTeX.

    My solution:



    First, the spaces that beero1000 listed all have this metric:
    ds^2 = dr^2 + S(r)^2 d\theta^2
    where s is the distance, r and θ are polar coordinates, and
     S(r) = \left\{ \begin{array}{l l} R \sin(r/R) & \text{elliptic} \\ r & \text{flat} \\ R \sinh(r/R) & \text{hyperbolic} \end{array} \right.

    he integral for distance is
     \displaystyle s = \int \sqrt{c^2 + S(c\theta)^2} \,d\theta

    In the flat case,
     \displaystyle s = c \int \sqrt{1 + \theta^2} \,d\theta = \frac{c}{2} \left( \theta \sqrt{1 + \theta^2} + \text{arcsinh} \theta \right)

    The non-flat cases require Jacobi elliptic integrals, and those are rather complicated.



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