The Math Thread

[beero1000]

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.

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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!

 

[Bomb#20]

The "sneaky" approach:

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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.

...
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?

 

[Kharakov]

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.

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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.

 

[beero1000]

That's a lot smarter than what I did.

...
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.

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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 \(m \geq n > 0\).

 

[Artemus]

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
\(\displaystyle = \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} x^{- 2}\, dx =\lim_{n\to\infty} (-x)^{-1} |_{x = 1/(n+1)}^{1/n} = \lim_{n\to\infty}-[n - (n+1)] = 1. \)

 

[lpetrich]

I came rather later to this party, so I've ended up duplicating others' solutions.

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:
e_n = sub of n x's all distinct -- these appear in the polynomial's coefficients
p_n = 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.

 

[lpetrich]

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 2^3n-1 + 5(3ⁿ) divisible by 11?

Consider the function f = 2^6n-1 + 5*3²ⁿ

It satisfies recurrence relation f(n+2) -73*f(n+1) + 576*f = 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.

 

[beero1000]

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...

 

[Kharakov]

I solved the |c|=0 case. I'll leave the |c| > 0 case for the rest of you.

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

 

[lpetrich]

First, LaTeX/Mathematics - Wikibooks, open books for an open world goes into detail about LaTeX.

My solution:

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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.

 

[lpetrich]

The elliptic version

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It uses Jacobi elliptic functions and integrals.
\( \displaystyle s = \int \sqrt{ c^2 + R^2 \sin^2 ((c/R) \theta) } \, d\theta = c E(m,\phi) ,\ m = - (R/c)^2 ,\ \phi = (c/R)\theta\)
E is the second kind of elliptic integral.

One can get the hyperbolic case by analytic continuation.

 

[beero1000]

Classes just ended and I've suddenly got a lot more time...

1. Find the largest area rectangle inside a semicircle.
2. Find the largest area triangle inside a semicircle.
3. Find the largest area semicircle inside a unit square.
4. Find the 100th digit after the decimal of \((1 + \sqrt{2})^{2000}\)
5. The largest known prime is p = 2^57,885,161 − 1. Find the last (least significant) digit of p. Find the first (most significant) digit of p.

... and one of my favorites:

1. Given 10 points in the plane, can they always be covered by (at most) 10 disjoint unit disks?

 

[Bomb#20]

5. The largest known prime is p = 2^57,885,161 − 1. Find the last (least significant) digit of p. Find the first (most significant) digit of p.

Too easy. The first digit is 1. The last digit is f.

 

[George S]

• The largest known prime is p = 2^57,885,161 − 1. Find the last (least significant) digit of p. Find the first (most significant) digit of p.

Too easy. The first digit is 1. The last digit is f.

Easier: Both the first and last digit are 1 (in base 2). 2^x-1 = 1, 11, 111, 1111, 11111...1111

 

[beero1000]

 

[Kharakov]

Too easy. The first digit is 1. The last digit is f.

Easier: Both the first and last digit are 1 (in base 2). 2^x-1 = 1, 11, 111, 1111, 11111...1111

Keeping with base 2....

\( \sqrt{ 10 - \sqrt{10+\sqrt{10+\sqrt{10+\sqrt{10+\sqrt{10 + \dots }}}}}}\)

Produces a specific pattern of base 2 digits after a whole bunch of zeros.
The number of times one must shift the pattern to the left, past the decimal point a bit, to match a familiar number, matches the number of 0s in the equation.
If the symbol for zero was any rounder, my reasoning would be even more circular.

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I can't Pell if I should write out a formula for a specific number to multiply by the square root of 2, for the 2^tooth question. 2 punny? I assume a simple entertaining trick, because it's you. Ohh.. hahhaa.. nm. Awesome.

I believe the answer to the question "Is there no trick?" is no, in German.


1 and 2 must be r^2. 3. I cut the pizza 4 times, and took one slice.

 

[lpetrich]

Classes just ended and I've suddenly got a lot more time...

Thanx.

Find the largest area rectangle inside a semicircle.

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The rectangle must fit against the straight side, and it must extend to the circular side. Will use unit radius for convenience. For w along the straight side and h perpendicular to it, we must minimize area A
\( A = 2wh ,\ w^2 + h^2 = 1 \)
Easy to do with Lagrange multipliers:
\( 2h - 2\lambda w = 0 ,\ 2w - 2\lambda h = 0\)
Solution: w = h = 1/sqrt(2), λ = 1.

Find the largest area triangle inside a semicircle.

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Two of the points are at the line-circle intersections, while the third one is on the circle part, at along-line offset from center w and perpendicular offset h. The triangle's area is (1/2)*2*h = h. So the maximum-area triangle is a right triangle with the right-angle vertex on a line perpendicular from the center.

Find the largest area semicircle inside a unit square.

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Let the square's vertices be A11 - A12 - A22 - A21 and back to A11. The straight side intersects A11 - A12 at B1 and A11 - A21 at B2, and the circular side touches A12 - A22 at C1 and A21 - A22 at C2. Let the circle's radius be r and its center be O.

Let angle B1 - O - C1 be a. Then B2 - O - C2 will be 90d - a. Projecting from C2 to O to A11 - A12 gives total length r*(1 + sin(a)), and doing likewise from C1 to O to A11 - A21 gives total length r*(1 + cos(a)). This means sin(a) = cos(a) or a = 45d. That also yields radius r = 2 - sqrt(2).

Find the 100th digit after the decimal of \((1 + \sqrt{2})^{2000}\)

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This number is approximatly 10^765.551 and its reciprocal 10^-765.551. However, this number can be expressed in form a + b, where a is an integer and b an integer times sqrt(2). Its reciprocal is likewise a - b. Thus, b = a - 10^-765.551 and a + b = 2a - 10^-765.551. Since a is an integer, the number must have about 765 decimal digits that are 9. Thus, the 100th digit is 9.

The largest known prime is p = 2^57,885,161− 1. Find the last (least significant) digit of p. Find the first (most significant) digit of p.

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When one takes powers of 2, the digits cycle as follows:
4p: 6
4p+1: 2
4p+2: 4
4p+3: 8

Since the exponent has form 4p+1, the least significant digit thus equals 2 - 1 = 1.

I found the most significant digit by brute force using Mathematica. It is 5.

Given 10 points in the plane, can they always be covered by (at most) 10 disjoint unit disks?

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[1101.3468] Packing-constrained point coverings considers what number of points have some configuration that cannot be covered by that number of unit disks. The number is >= 11 and <= 55. So it's always possible for 10 points.

 

[beero1000]

The largest known prime is p = 2^57,885,161− 1. Find the last (least significant) digit of p. Find the first (most significant) digit of p.

Show

When one takes powers of 2, the digits cycle as follows:
4p: 6
4p+1: 2
4p+2: 4
4p+3: 8

Since the exponent has form 4p+1, the least significant digit thus equals 2 - 1 = 1.

I found the most significant digit by brute force using Mathematica. It is 5.

For the most significant digit:

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Look at the fractional part of 57885161 log₁₀2...

Given 10 points in the plane, can they always be covered by (at most) 10 disjoint unit disks?

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[1101.3468] Packing-constrained point coverings considers what number of points have some configuration that cannot be covered by that number of unit disks. The number is >= 11 and <= 55. So it's always possible for 10 points.

The current best known bounds are 13 and 45, respectively . However, there is a particularly nice argument for the n ≤ 10 case.

 

[beero1000]

Covering 10 points with disjoint circles.

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Take a regular packing of circles, and center it randomly in the plane. Note that the proportion of the plane that is covered by the circles is \(\pi/2\sqrt{3} \approx 0.9069\), so the expected number of points covered by the circles is about 9.069. Since 9.069 > 9, there must exist packing placements that cover all 10 points.

Isn't the probabilistic method wonderful?

 

[beero1000]

I recently found http://www.brilliant.org. It is a training site for high school math competitions. You work up in levels to get to harder problems.

Fun!