The Math Thread

[beero1000]

For those interested (or bored) enough, a new thread to pose/solve math problems.

1. 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).
2. 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.
3. Is (14n+3)/(21n + 4) a fraction in simplest form for every n?
4. Two unit circles pass through each other's center. Find the area of their intersection. Bonus points: Same question for 3 intersecting circles.

 

[ApostateAbe]

I have the solution to #1.

x^5 + 2*x^4 + 3*x^3 + 4*x^2 + 5*x + 6 = 0

http://www.mathportal.org/calculators/solving-equations/polynomial-equation-solver.php

a = -1.4918
b = 0.55169+1.25335*i
c = 0.55169-1.25335*i
d = -0.80579+1.2229*i
e = -0.80579-1.2229*i

a^2+b^2+c^2+d^2+e^2=

(-1.4918)^2+(0.55169+1.25335*i)^2+(0.55169-1.25335*i)^2+(-0.80579+1.2229*i)^2+(-0.80579-1.2229*i)^2=

(https://www.google.com/search?q=(-1...j0.330j0j7&sourceid=chrome&es_sm=122&ie=UTF-8)

-1.9999552646

approximately equals (could very well be exact solution)

-2

Google has made me an expert mathematician.

I will solve the others later.

 

[beero1000]

I have the solution to #1.

x^5 + 2*x^4 + 3*x^3 + 4*x^2 + 5*x + 6 = 0

http://www.mathportal.org/calculators/solving-equations/polynomial-equation-solver.php

a = -1.4918
b = 0.55169+1.25335*i
c = 0.55169-1.25335*i
d = -0.80579+1.2229*i
e = -0.80579-1.2229*i

a^2+b^2+c^2+d^2+e^2=

(-1.4918)^2+(0.55169+1.25335*i)^2+(0.55169-1.25335*i)^2+(-0.80579+1.2229*i)^2+(-0.80579-1.2229*i)^2=

(https://www.google.com/search?q=(-1...j0.330j0j7&sourceid=chrome&es_sm=122&ie=UTF-8)

-1.9999552646

approximately equals (could very well be exact solution)

-2

Google has made me an expert mathematician.

I will solve the others later.

The exact value is indeed -2. The main question is ... can you prove it?

 

[ApostateAbe]

I am an empiricist and I care not for damn proofs!

 

[beero1000]

I am an empiricist and I care not for damn proofs!

Well then, I guess you could empirically determine the sum of the squares of the roots of the polynomial equation a₅x⁵+a₄x⁴+a₃x³+a₂x²+a₁x+a₀=0.

 

[Bomb#20]

2. 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.

Area = 42.627
Radius = 3.957

Or in closed form...

Show

Area = 20 + 16 * sqrt(2)
Radius = sqrt(10 + 4 * sqrt(2))

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

Yes. Argument...

Show

gcd(14n+3, 21n + 4)
= gcd(14n+3, 7n + 1)
= gcd(0n+1, 7n + 1)
= 1

 

[Bomb#20]

4. Two unit circles pass through each other's center. Find the area of their intersection. Bonus points: Same question for 3 intersecting circles.

Two: 1.228
Three: 0.7048

Or in closed form...

Show

Inscribe a hexagon in a unit circle. It is composed of six unit-sided equilateral triangles. Let A be the area of one of them; let B be the area of one of the six slivers outside the hexagon but inside the circle. Then

A = sqrt(3) / 4
B = pi / 6 - A

By inspection, the intersection of two unit circles is 2A + 4B; the intersection of three unit circles is A + 3B.

 

[Bomb#20]

I am an empiricist and I care not for damn proofs!

Well then, I guess you could empirically determine the sum of the squares of the roots of the polynomial equation a₅x⁵+a₄x⁴+a₃x³+a₂x²+a₁x+a₀=0.

Why, can you do it analytically?

ETA: Hey, Newton's Identities! Cool! And I'd thought Galois's proof meant we were screwed for the general quintic equation.

 

[beero1000]

Well then, I guess you could empirically determine the sum of the squares of the roots of the polynomial equation a₅x⁵+a₄x⁴+a₃x³+a₂x²+a₁x+a₀=0.

Why, can you do it analytically?

ETA: Hey, Newton's Identities! Cool! And I'd thought Galois's proof meant we were screwed for the general quintic equation.

Fun, right? (AKA  Vieta's formulas)

 

[beero1000]

1. If x,y,z are positive real numbers with xyz = 1, show that x²+y²+z² ≤ x³+y³+z³
2. If n is a positive even number, is 2^3n-1 + 5(3ⁿ) divisible by 11?
3. What is \(\displaystyle \lim_{n\to\infty} \int_{1/(n+1)}^{1/n} \frac{\sin x}{x^3} dx\)?
4. Find the surface area and volume of a horn torus (a circle rotated about one of its tangents).

 

[ryan]

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.

What did I do wrong?

4b + 4c = 2pi, where b is the arc of the 2 unit long line segment, and c is the other.

c = pi/2 - b

r = 2/sin(c)

r = 1/sin(b)

2sin(b) = sin(c)

2sin(b) = sin(pi/2 - b) = cos(b)

b = tan^(-1)(1/2)

r = 1/sin(b) = 2.236...

 

[Bomb#20]

2. If n is a positive even number, is 2^3n-1 + 5(3ⁿ) divisible by 11?

Yes.

Show

Change variables and write the formula mod 11. Let n = 2m+2, m a non-negative integer. So

2^3n-1 + 5(3ⁿ) = 2^6m+5 + 5(3^2m+2)
= 32(2⁶)^m + 5(9)(3²)^m
= -1(9^m) + 1(9^m) (mod 11)
= 0 (mod 11)

 

[ryan]

2. If n is a positive even number, is 2^3n-1 + 5(3ⁿ) divisible by 11?

Yes.

Show

Change variables and write the formula mod 11. Let n = 2m+2, m a non-negative integer. So

2^3n-1 + 5(3ⁿ) = 2^6m+5 + 5(3^2m+2)
= 32(2⁶)^m + 5(9)(3²)^m
= -1(9^m) + 1(9^m) (mod 11)
= 0 (mod 11)

I used induction.

n = 2m, where m is any natural number.

Assume, 2^{6m - 1} + 5(3^2m) = 11q, where q is another natural number.

m = 1: 77 = 11q

m + 1: 2^{6m + 5} + 5(3^{2m + 2}) = 2^{6m - 1} + 5(3^{2m + 2}) + 11

 

[Bomb#20]

I used induction.

n = 2m, where m is any natural number.

Assume, 2^{6m - 1} + 5(3^2m) = 11q, where q is another natural number.

m = 1: 77 = 11q

m + 1: 2^{6m + 5} + 5(3^{2m + 2}) = 2^{6m - 1} + 5(3^{2m + 2}) + 11

You must have dropped a term -- for m=1, that equation works out to 2453 = 448.

(Which isn't to say that induction couldn't work.)

What did I do wrong? ... sin(c) ...

Tried to use trig!

(Which isn't to say that trig couldn't work. But I got away with using the Pythagorean Theorem and a couple of pages of algebra.)

 

[ryan]

You must have dropped a term -- for m=1, that equation works out to 2453 = 448.

(Which isn't to say that induction couldn't work.)

thanks

Let/assume 2^{6m - 1} + 5(3^2m) = 11q.

m + 1: 2⁶(2^{6m - 1}) + 5(3^{2m + 2}

= 2⁶(2^{6m - 1}) + 3²*5(3^2m)

= 2⁶(2^{6m - 1}) + 9(11q - 2^{6m - 1})

= 2⁶(2^{6m - 1}) + 9(11q) - 9(2^{6m - 1})

= 2^{6m - 1}(2⁶ - 9) + 9(11q)

= 2^{6m - 1}(55) + 9(11q)

= 11((2^{6m - 1})(5) + 9q)

What did I do wrong? ... sin(c) ...

Tried to use trig!

(Which isn't to say that trig couldn't work. But I got away with using the Pythagorean Theorem and a couple of pages of algebra.)

But where did I go wrong? Every step seems correct.

 

[Bomb#20]

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.

Show

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.

 

[beero1000]

You must have dropped a term -- for m=1, that equation works out to 2453 = 448.

(Which isn't to say that induction couldn't work.)

Induction will work, as long as he is careful with his algebra. As written, there are errors.

What did I do wrong? ... sin(c) ...

Tried to use trig!

(Which isn't to say that trig couldn't work. But I got away with using the Pythagorean Theorem and a couple of pages of algebra.)

Trig will work, if you're careful, but I agree it's not the best approach.

The "sneaky" approach:

Show

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 \(\sqrt{(2+\sqrt{2})^2 + 2^2} = \sqrt{10 + 4\sqrt{2}}\)

 

[ryan]

Trig will work, if you're careful, but I agree it's not the best approach.

I am so frustrated because each step seems right. Please - I beg you - take me out of my misery!

 

[beero1000]

Trig will work, if you're careful, but I agree it's not the best approach.

I am so frustrated because each step seems right. Please - I beg you - take me out of my misery!

r = 2/sin(c/2) not r = 2/sin(c)

 

[ryan]

I am so frustrated because each step seems right. Please - I beg you - take me out of my misery!

r = 2/sin(c/2) not r = 2/sin(c)

Oh for the love of ...

Thanks!!!!!!