Micromass' big October challenge

[micromass]

problem 4 highshool:
x coordinate = $-1+\frac{1}{2}-\frac{1}{4}... = -1 + \frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}...) = -1 + \frac{log(2)}{2}$
y coordinate = $1-\frac{1}{3}+\frac{1}{5}..=\frac{\pi }{4}$
converges to the point $\left ( -1+\frac{log(2)}{2},\frac{\pi }{4} \right )$

Read the problem more carefully. The lengths are defined recursive.

 

[MAGNIBORO]

Read the problem more carefully. The lengths are defined recursive.

my mistake,
x coordinate: series of cos is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}...$ so $-1+\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}...=-cos(1)$
y coordinate: series of sin is $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$ so $1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}...=sin(1)$

for problem 9:
$I_{n}= \int_{0}^{1}x^n\sqrt{1-x}\, dx = \beta (n+1,\frac{3}{2})=\frac{n! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{3}{2})!}$
and $I_{n-1}= \frac{(n-1)! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{1}{2})!}$
so
$I_{n-1} \: \frac{2n}{2n+3}= \frac{(n-1)! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{1}{2})!} \: \frac{n}{n+\frac{3} {2}} = \frac{n! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{3}{2})!} = I_{n}$

 

[MAGNIBORO]

for high school problem 5 (very entertaining) :
$A(1,n) = A(0,A(1,n-1)) = 1 + A(1,n-1)$
$1 + A(1,n-1) = 2 + A(1,n-2)$
$...$
$A(1,n) = n + A(1,0) = n+2$

$A(2,0) = A(1,1) = 1 + 2 = 3$
$A(2,n) = A(1,A(2,n-1)) = A(2,n-1) + 2$
$2 + A(2,n-1) = 4 + A(2,n-2)$
$...$
$A(2,n) = 2n + 3$

$A(3,0) = A(2,1) = 5$
$A(3,1) = A(2,A(3,0)) = 2 A(3,0) + 3 = 13$
$A(3,2) = 2 A(3,1) +3 = 29$
$A(3,3) = 2 A(3,2) + 3 = 61$
now note that
$A(3,1) - A(3,0) = 8$
$A(3,2) - A(3,1) = 16$
$A(3,3) - A(3,2) = 32$
so
$A(3,n+1) = 2 A(3,n) + 3 = A(3,n) + 2^{n+3}$
$A(3,n) = 2^{n+3} - 3$

 

[MAGNIBORO]

Problem 10:
$$\int_{-1}^{1}\frac{dx}{x^2+1}=\frac{\pi }{2}$$
$$\int_{-1}^{1}\frac{e^{x}+1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{2}$$
$$\int_{-1}^{1}\frac{e^{x}}{(x^2+1)(e^{x}+1)}+\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{2}$$
we can assume that the 2 integrals are equal, and this is true if
$$\int_{-1}^{1}\frac{1-e^{x}}{(x^2+1)(e^{-x}+1)}=0$$
That would be true if $\frac{1-e^{x}}{(x^2+1)(e^{-x}+1)}$ is a odd function
$$f(-x)=-f(x)$$
$$\frac{1-e^{-x}}{(x^2+1)(e^{-x}+1)}=-\frac{1-e^{x}}{(x^2+1)(e^{x}+1)}$$
$$\frac{-1+e^{x}}{(x^2+1)(e^{x}+1)}=-\frac{1-e^{x}}{(x^2+1)(e^{x}+1)}$$
so
$$\int_{-1}^{1}\frac{e^{x}}{(x^2+1)(e^{x}+1)}=\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}$$
$$\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{4}$$
this problem is very good, and It has a very beautiful relationship:
$$\int_{-1}^{1}\frac{1}{(x^2+1)(e^{-x}+1)}+\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{2}$$Problem 6:
$$I=\int_{0}^{\frac{\pi }{2}}\frac{sin(x)cos(x)}{x+1}=\frac{1}{2}\left ( \left[ \frac{sin^2(x)}{x+1} \right]_{0}^{\frac{\pi }{2}}+\int_{0}^{\frac{\pi }{2}}\frac{sin^2(x)}{(x+1)^2}
\right )=\frac{1}{2}\left ( \frac{2}{\pi+2 }+ \frac{1}{2}\left ( \int_{0}^{\frac{\pi }{2}}\frac{1}{(x+1)^2}-\int_{0}^{\frac{\pi }{2}}\frac{cos(2x)}{(x+1)^2} \right )\right )$$
$$=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{cos(2x)}{(x+1)^2}$$
$$=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{4}\int_{1}^{1+\frac{\pi}{2}}\frac{cos(2u-2)}{u^2}$$
$$=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{4}\left (cos(2)\int_{1}^{1+\frac{\pi}{2}}\frac{cos(2u)}{u^2}+sin(2)\int_{1}^{1+\frac{\pi}{2}}\frac{sin(2u)}{u^2} \right )$$
$$I=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{2}\left (cos(2)\int_{2}^{2+\pi}\frac{cos(u)}{u^2}+sin(2)\int_{2}^{2+\pi}\frac{sin(u)}{u^2} \right )$$
introduce M
$$I=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{M}{2}$$

now the other integral.

$$J=\int_{0}^{\pi }\frac{cos(x)}{(x+2)^2}=\int_{2}^{\pi +2}\frac{cos(u-2)}{u^2}=cos(2)\int_{2}^{2+\pi}\frac{cos(u)}{u^2}+sin(2)\int_{2}^{2+\pi}\frac{sin(u)}{u^2}$$
$$J=M$$
$$I=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{J}{2}$$

 

[zinq]

In Problem 2) ("Let p≠0 be a real number ...") it is interesting to look at what happens as p → 0. The problem is still meaningful in that case!

 

[Stella.Physics]

so what about the answer to the birds on wire problem? Really eager to find out what the answer and proof is :)

 

[Erland]

A clue to advanced problem 3 would also be appreciated...

 

[parshyaa]

CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

1) let A,B,C,D be a complex numbers with length 1. Prove that if A+B+C+D=0, then these four numbers form a rectangle.

2) On an arbitrary triangle, we produce on each side an equilateral triangle. Prove that the centroids of these three triangles forms an equilateral triangle

• Are they unsolved

 

[parshyaa]

CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

1) let A,B,C,D be a complex numbers with length 1. Prove that if A+B+C+D=0, then these four numbers form a rectangle.

2) On an arbitrary triangle, we produce on each side an equilateral triangle. Prove that the centroids of these three triangles forms an equilateral triangle

• Are they unsolved

Hey i got the answer to the second of these questions, please tell me it is solved or unsloved

 

[parshyaa]

 

[mfb]

If they are not marked as solved (and if the last 2-3 posts don't cover them), then no one posted a solution yet.

 

[parshyaa]

If they are not marked as solved (and if the last 2-3 posts don't cover them), then no one posted a solution yet.

It means they are unsloved, thanks

 

[parshyaa]

@micromass

 

[parshyaa]

Where is micromass

 

[mfb]

Currently too busy to maintain the challenges threads.

 

[SWest1234]

I have a solution to high school Q8.
We can factor the polynomial 5x^3+6x^2+5x+4 as(x+1)(5x^2+x+4).
5x^2+x+4 >x+1 for all x. so let x+1=p^a and 5x^2+x+4=p^b where b>a. Divided, we obtain (5x^2+x+4)/(x+1)=p^b-a, which is an integer. This can be written as 5x-4+8/(x+1). For this to be an integer, we must have x+1 dividing 8, which means the only solution is base 7, in which case we get 56547=204810 which equals 2^11.

 

[mfb]

Correct.

 

[EpidermalOblivion]

Question 1 for high school and first-year university:

Would this be a counterexample?

Θ = any non-right angle
$$\\<br /> A = \cos \Theta + i\sin \Theta \\<br /> B = 0 + i \\<br /> C = -\cos \Theta - i\sin \Theta \\<br /> D = 0 - i$$

$A + B + C + D = 0$, but a rhombus is formed instead of a rectangle.
I probably simply misinterpreted the question or do not fully understand it.

 

[EpidermalOblivion]

Question 1 for high school and first-year university:

Would this be a counterexample?

Θ = any non-right angle
$$\\<br /> A = \cos \Theta + i\sin \Theta \\<br /> B = 0 + i \\<br /> C = -\cos \Theta - i\sin \Theta \\<br /> D = 0 - i$$

$A + B + C + D = 0$, but a rhombus is formed instead of a rectangle.
I probably simply misinterpreted the question or do not fully understand it.

Nevermind, I realized that this is wrong; it would still be a rectangle. This is my proof for the original problem:

Two complex numbers of length 1 can be represented by a line segment spanning the diameter of a unit circle. Because these two lines are of opposite direction but equal length, their sum as complex numbers is 0. By using 4 complex numbers, two perpendicular lines can be formed creating a square. By altering the angle of the lines, one may form an infinite number of distinct rectangles. Any individual complex number in a line may not be altered independently of the complex number forming the other half of the line as this would cause the sum of the two to no longer equal 0. This holds as long as all of the complex numbers are distinct, otherwise the two diameter-length line segments could be merged into one line.

 

[mfb]

Two complex numbers of length 1 can be represented by a line segment spanning the diameter of a unit circle. Because these two lines are of opposite direction but equal length, their sum as complex numbers is 0.

Not all pairs of complex numbers of length 1 do that. Consider a=1 and b=i. Both have length 1 but the line segment doesn't go through zero and they don't sum to zero. Which is not necessary, because only all 4 values together have to have a sum of zero.

It is possible to show that the 4 values have to consist of two pairs as you describe them, but that part is not trivial.

 

[EpidermalOblivion]

Not all pairs of complex numbers of length 1 do that. Consider a=1 and b=i. Both have length 1 but the line segment doesn't go through zero and they don't sum to zero. Which is not necessary, because only all 4 values together have to have a sum of zero.

It is possible to show that the 4 values have to consist of two pairs as you describe them, but that part is not trivial.

The sum of two complex numbers of length 1 will only be zero if the two numbers are pointed in the opposite direction. This is because the real and imaginary parts of the two numbers will be additive inverses of each other. This cannot be done with, for example, 3 complex numbers of the same length, as the first 2 would cancel and the last would take the sum back away from zero. In general, the sum of complex numbers can be zero only with an even number of complex numbers. If the complex numbers must be distinct, this can only occur with 2 complex numbers. Therefore, 4 distinct complex numbers summing to zero must be represented by two distinct lines with a length of 2, forming a rectangle.

 

[Charles Link]

The sum of two complex numbers of length 1 will only be zero if the two numbers are pointed in the opposite direction. This is because the real and imaginary parts of the two numbers will be additive inverses of each other. This cannot be done with, for example, 3 complex numbers of the same length, as the first 2 would cancel and the last would take the sum back away from zero. In general, the sum of complex numbers can be zero only with an even number of complex numbers. If the complex numbers must be distinct, this can only occur with 2 complex numbers. Therefore, 4 distinct complex numbers summing to zero must be represented by two distinct lines with a length of 2, forming a rectangle.

$1, exp^{i \frac{2 \pi}{3}}, \, and \, exp^{i \frac{4 \pi}{3}}$ all have magnitude 1 and their vector sum is zero.

 

[mfb]

This cannot be done with, for example, 3 complex numbers of the same length, as the first 2 would cancel and the last would take the sum back away from zero. In general, the sum of complex numbers can be zero only with an even number of complex numbers.

Counterexample:
$1$
$-\frac{1}{2} + \frac{\sqrt{3}}{2} i$
$-\frac{1}{2} - \frac{\sqrt{3}}{2} i$
Edit: Charles Link was faster.

Therefore, 4 distinct complex numbers summing to zero must be represented by two distinct lines with a length of 2, forming a rectangle.

The "therefore" has nothing that would lead to such a conclusion.Looking at pairs of complex numbers can be useful, but you have to allow their sum to be non-zero.

 

[EpidermalOblivion]

Counterexample:
$1$
$-\frac{1}{2} + \frac{\sqrt{3}}{2} i$
$-\frac{1}{2} - \frac{\sqrt{3}}{2} i$
Edit: Charles Link was faster.
The "therefore" has nothing that would lead to such a conclusion.Looking at pairs of complex numbers can be useful, but you have to allow their sum to be non-zero.

Thank you for pointing this out, I did not consider that. I have revised my proof:

Consider these 4 complex numbers forming 2 perpendicular lines of length two:
$$\\<br /> A = 1<br /> \\<br /> B = i<br /> \\<br /> C = -1<br /> \\<br /> D = -i$$
These 4 complex numbers sum to zero. They also form a rectangle (specifically, a square). From here, one may prove that the complex numbers may only be manipulated such that one of the two lines is rotated, which would only alter the dimensions of the rectangle formed.
The sum of two complex numbers of length 1 will only be zero if the two numbers are pointed in opposite directions. This is because the real and imaginary parts of the two numbers will be additive inverses of each other. This means for each pair of complex numbers forming a line ($A$ and $C$, $B$ and $D$), one may only rotate the two complex numbers jointly as a single line.
If one were to alter the orientation of only one complex number, the sum would no longer be zero. Similarly, if one were to rotate two perpendicular complex numbers (such as $A$ and $B$) together, the sum would also no longer be zero as both lines would be broken. If one were to rotate 3 complex numbers at once, this would be the same as rotating one number (the transformation is the same relative to the one complex number not being altered). This would cause the sum to be nonzero and also cause the shape to no longer be a rectangle. If one were to rotate 4 complex numbers at once, the effect would be the same as rotating each of the 2 lines by the same amount. The sum would still be zero and the rectangle would simply be rotated.
All transformations upon these lines that would cause the lines to no longer form a rectangle can not be performed as long as the sum is zero. Therefore, 4 distinct complex numbers summing to zero must form a rectangle.

 

[mfb]

If one were to alter the orientation of only one complex number, the sum would no longer be zero. Similarly, if one were to rotate two perpendicular complex numbers (such as A and B) together, the sum would also no longer be zero.

What if you alter three or four numbers at the same time? You didn't cover these cases.

 

[EpidermalOblivion]

What if you alter three or four numbers at the same time? You didn't cover these cases.

Altering 3 numbers does not work. This is the same as altering one of the two lines (which works) and then altering another line independently of its counterpart, which will cause the sum to be nonzero for reasons previously explained. The fact that this does not work supports the conjecture as if this was possible the 4 numbers would no longer form a rectangle.
Altering 4 numbers works because it is the same as altering one of the two lines and then altering the other line. This is the equivalent of simply rotating the entire rectangle.

 

[mfb]

Altering 3 numbers does not work. This is the same as altering one of the two lines (which works) and then altering another line independently of its counterpart, which will cause the sum to be nonzero for reasons previously explained.

Only if you assume two numbers stay on opposite sides.

You keep using the result you want to prove in the proof.

As an example, I could use your arguments to "show" that 6 complex unit-length numbers would need three pairs that sum to zero each. But they don't have to. Your proof cannot be right.

 

[EpidermalOblivion]

Only if you assume two numbers stay on opposite sides.

I have considered this and added it along with the cases for altering 3 and 4 numbers in my proof above (post #114).

As an example, I could use your arguments to "show" that 6 complex unit-length numbers would need three pairs that sum to zero each. But they don't have to. Your proof cannot be right.

I do not see how this could be done. My proof involves starting with a condition in which the complex numbers sum to zero and form a rectangle and then attempting to prove that one could not alter the complex numbers such that the resulting sum is zero and does not form a rectangle. It does not apply to cases in which there are 6 complex numbers.

 

[mfb]

I do not see how this could be done.

That is exactly the problem in all your posts. You assume that you need those pairs, and then you conclude that you need those pairs.

An example for 6 numbers without pairs summing to zero: $e^{0}$ and $e^{i 2\pi/3}$ and $e^{i 4\pi/3}$ and $e^{i}$ and $e^{i (1+2\pi/3)}$ and $e^{i(1+ 4\pi/3)}$

If one were to rotate 3 complex numbers at once, this would be the same as rotating one number (the transformation is the same relative to the one complex number not being altered).

It is not the same.

 

[EpidermalOblivion]

Question 1 for high school and first-year university:
I have once again revised my proof for this problem. I assume that each complex number must be distinct, otherwise, one could arrange all of the numbers in a line of length 2, which would not form a rectangle yet sum to zero.
By rearranging $A + B + C + D = 0$, one may derive that $A + B = -C + -D$. This means that for each two complex numbers, their sum must be mirrored by the sum of the other two complex numbers. This is depicted in the diagram below, where the cyan line must mirror the yellow line and the pink line must mirror the green line.

For any two unit-length complex numbers that are distinct and are not opposites of each other, the origin, their sum, and the two numbers will form a parallelogram where each side has a length of 1. This means that the line drawn between the two complex numbers will be perpendicular to the line drawn between the origin and their sum. Therefore, the sum of the two unit-length complex numbers will be an angular bisector of the two.

Imagine trying to rotate one sum independently. The sum must remain the angular bisector of the two numbers that it is the sum of. In order to do this, one must rotate both of these numbers by the same amount that one rotates the sum. In doing this, one will cause the two other sums perpendicular to the sum (in the case of $A+B$, these would be $A+D$ and $B+C$) to no longer be angular bisectors. Any rotation of two or three sums may be broken into moving each sum individually and therefore will not work either.
Decreasing the magnitude of one sum independently increases the angle between the two numbers that it is the sum of. Analogously, increasing the magnitude of one sum independently decreases the angle between the two numbers that it is the sum of. This is impossible to do without causing 2 other sums to no longer be angular bisectors. Scaling three sums together by the same amount may be interpreted as scaling two sums and then scaling one independently (which does not work as previously explained).

This limits one to only rotating all of the sums together or scaling the magnitudes of two opposite sums by the same amount. Rotating all four sums together would rotate all four numbers together, which would not affect whether or not the numbers form a rectangle. For the four complex numbers to form a rectangle, the angle between any two complex numbers must be equal to the angle between the other two. For example, the angle between $A$ and $B$ must be equal to the angle between $C$ and $D$. By altering the magnitude of two opposite sides together, one simply increases/decreases the angles within each of the two pairs of numbers by the same amount. From the arrangement in the diagram above, any transformation will preserve the equivalence between the angle of any two complex numbers and the other two. Because this condition will always be satisfied, the four complex numbers must form a rectangle.