Mathematics marathon

[Keroma12]

Spoiler

How did you derive the first equation?

Spoiler

Draw the diagram, then the area of the inner square will be the area of the whole square (1) minus the 4 larger triangles [(x)(1)/2] plus the 4 smaller triangles [(xcos\( \theta \))(xsin\( \theta \))/2]
So y = 1-2x+2x^2sin\( \theta \)cos\( \theta \)
And from the larger triangle, sin\( \theta \) = \( \frac{x}{\sqrt{x^2+1}} \) and cos\( \theta \) = \( \frac{1}{\sqrt{x^2+1}} \)
Plug those in and then simplify.
This is what my first instinct was to do. Is there a better way? And am I even right?

 

[qqwref]

Spoiler

Draw the diagram, then the area of the inner square will be the area of the whole square (1) minus the 4 larger triangles [(x)(1)/2] plus the 4 smaller triangles [(xcos\( \theta \))(xsin\( \theta \))/2]
So y = 1-2x+2x^2sin\( \theta \)cos\( \theta \)
And from the larger triangle, sin\( \theta \) = \( \frac{x}{\sqrt{x^2+1}} \) and cos\( \theta \) = \( \frac{1}{\sqrt{x^2+1}} \)
Plug those in and then simplify.

Spoiler

Sure, that works fine. A possibly easier way is:
y = 1 - 4*(area of small triangle + trapezoid, a shape which is itself a right triangle)
= \( 1 - 4*(\frac{1}{2} \sin(\theta) \cos(\theta)) \) (those are the two non-hypotenuse sides of the triangle)
= \( 1 - 2*(\frac{x}{\sqrt{x^2+1}} * \frac{1}{\sqrt{x^2+1}}) \)
= \( 1 - \frac{2x}{x^2+1} \)
= \( \frac{x^2 - 2x + 1}{x^2 + 1} \)

.

 

[Rinfiyks]

A problem with no nasty calculus or algebra:

Spoiler

It's a unit square. B and F are midpoints. Find the area of the shaded region.

 

[jerry]

Hi, I'm doing As Level Maths at the minute, and I have a fair interest in maths, and I would like to pose this question, its just something I noticed when I was bored in a physics lesson:

Spoiler

Why is it that:
sin cos cos sin (n)
sin cos cos sin cos cos cos sin (n)
sin cos cos sin cos cos cos sin cos cos cos cos sin (n)

Firstly, for any value of n, tend towards the same value, and secondly, why do they tend towards a more specific value as you work down the list (as the number of sin and cos functions are increased in that pattern)?

P.S Im not 100% sure of my answer...

I'm not sure if the following qualifies as a full/legitimate answer, but hopefully it'll give you a rough idea of what's happening. I encountered the same thing with my friend while messing with a calculator

Spoiler

I tried this too using cos(cos(cos(...x))))...). It tended towards about 0.7931. Compounding sine also tends to 0.7931, except it does so slower (I think). I learned the reason for this behavior with a google search. It turns out that 0.7931 is the "steady state" of the function, which is when the input equals the output.

I suggest you read about it. It's a bit interesting.
http://books.google.com/books?id=Oi...AA#v=onepage&q=cosine of cosine .7391&f=false

 

[maggot]

Given that \( f(x)=x \) is the solution to the DE
\( y'' - 2xy + 2y = 0 \)
determine a second linearly independent soln.

no need to break down, this is no elementary integration. ; w;

 

[vcuber13]

A problem with no nasty calculus or algebra:

Spoiler

It's a unit square. B and F are midpoints. Find the area of the shaded region.

it may be off by a bit because i only used like 3 decimals

Spoiler

\(
2.75 \sqrt{74}
\)
or about
\(
23.6564
\)

 

[keemy]

maggot: your problem is wrong y=x isn't a solution (unless you mean 2xy' for second term) in that case the solution is really nasty because it ends up being something like

Spoiler

\( x \int x^{-2} e^{x^2} dx \)

so pretty lame (someone can try for themselves it's pretty easy redux of order then solve the first order DE with an integrating factor or as a separable equation but then integral of that is as stated above).

and a problem (not that hard)

Spoiler

let \( a_0=c \) and \( a_1=k \) solve the recurrence \( a_{n+2} = \frac {-a_n}{2(n+2)} \) (if you want you can give separate solutions for even and odd)

 

[Quadrescence]

No calculus required.

Solve for \( x \):

\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)

where the pattern continues ad infinitum.

Look at this hint if you're stuck.

Spoiler

There is an awful lot of repetition going on here. If I said \( y = 2 - y \), then it'd also be true that \( y = 2 - (2 - y) \), and furthermore, \( y = 2 - (2 - (2 - y)) \), ad infinitum. Try to use this idea here.

 

[keemy]

quad's

Spoiler

2
\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( (4-x)^2 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 16-8x+x^2= 4 \)
\( (x-6)(x-2)=0 \)
6 dun werk so 2.

 

[Rinfiyks]

it may be off by a bit because i only used like 3 decimals

Spoiler

\(
2.75 \sqrt{74}
\)
or about
\(
23.6564
\)

Hmm, that's not the answer I have. I got a fraction. And it's not 1/3.

No calculus required.

Solve for \( x \):

\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)

where the pattern continues ad infinitum.
​

Spoiler

\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 4 = x + \sqrt{4} \)
\( x = 2 \)
I like problems like this.

A slightly harder problem:

Spoiler

I have 5 equally sized spheres. I put 4 of them on a flat surface. These 4 all touch the adjacent ones, like in the picture below, which is a view from above.

I then place the 5th sphere in the middle of the others. How high is it off the surface?

 

[Quadrescence]

quad's

Spoiler

2
\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( (4-x)^2 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 16-8x+x^2= 4 \)
\( (x-6)(x-2)=0 \)
6 dun werk so 2.

keemoop [and rinfiyks] are right, here is another way, basically equivalent to rinf's

Spoiler

\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 4 - x = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 4 - x = \sqrt{x + (4 - x)} \)
\( 4 - x = \sqrt{4} \)
\( 4 - x = 2 \)
\( x = 2 \)

 

[Quadrescence]

A slightly harder problem:

Spoiler

I have 5 equally sized spheres. I put 4 of them on a flat surface. These 4 all touch the adjacent ones, like in the picture below, which is a view from above.

I then place the 5th sphere in the middle of the others. How high is it off the surface?

Here's my initial stab at it.

Spoiler

Suppose the radii are \( r \). Then, connecting the spheres' centers, we have a square with every side length equal to \( 2r \) (because each sphere is tangent to one another, two centers of two such spheres will be no further than \( 2r \)). Since all faces (except the base) are equilateral triangles, this is called the first Johnson solid (\( J_1 \)). Not that that is relevant, but neato trivia.

Anyway to find the height, consider the distance to the center of the view given in the problem. The distance from the center of the image and any one of the four circle's centers is \( 2r/\sqrt{2} = r\sqrt{2} \) (why? construct a right triangle). Now we have the distance from the center of the pyramid's base, to one of its base vertexes. Now we have a right triangle if we travel along the edge of the pyramid to the apex. This is a right triangle, with base \( r\sqrt{2} \) and hypotenuse \( 2r \). Since \( \mathrm{base}^2+\mathrm{height}^2=\mathrm{hypotenuse}^2 \), we have

\( (r\sqrt{2})^2 + h^2 = (2r)^2 \)
\( 2r^2 + h^2 = 4r^2 \)
\( h^2 = 2r^2 \)
\( h = r\sqrt{2} \)

This is only the height from the plane which the centers of the four base spheres form, which is itself at a height of \( r \). So the center of the top sphere is at \( (1+\sqrt{2})r \). But if we want to know the distance the bottom of the sphere is from the ground, we just take off \( r \) from this, giving \( r\sqrt{2} \).

Not sure if this is correct, but it's my initial intuition.

 

[Quadrescence]

Spoiler

let \( a_0=c \) and \( a_1=k \) solve the recurrence \( a_{n+2} = \frac {-a_n}{2(n+2)} \) (if you want you can give separate solutions for even and odd)

Spoiler

Let's see here. My gut feeling is

\( a_n =
\begin{cases}
\displaystyle\left(-\frac{1}{2}\right)^{n/2}\frac{c}{n!!} & \text{even } n\\
\displaystyle\left(-\frac{1}{2}\right)^{(n-1)/2}\frac{k}{n!!} & \text{odd } n
\end{cases} \)
​

where \( n!! = n(n-2)(n-4)\cdots \) is the double factorial ending at \( 2 \) if \( n \) is even, \( 1 \) if odd. This can be "simplified" to

\( a_n = \left(-\frac{1}{2}\right)^{\lfloor n/2\rfloor}\frac{[(n+1)\bmod 2]c+(n\bmod 2)k}{n!!} \)
​

where \( n\bmod m \) is modular reduction.

 

[Quadrescence]

New problem: Define a rational function as one which is a quotient of polynomials. Let \( \Delta f(x) = f(x+1)-f(x) \), Show that no rational function \( f \) exists satisfying

\( \Delta f(x) = \frac{1}{x}. \)

If you're daring, show that

\( \Delta_k f(x) = \frac{1}{x} \)

​

has no rational solution, where \( \Delta_k f(x) = f(x+k)-f(x) \) for non-zero \( k\in\mathbb{Z} \).​

 

[cyoubx]

it may be off by a bit because i only used like 3 decimals

Spoiler

\(
2.75 \sqrt{74}
\)
or about
\(
23.6564
\)

How exactly did you get an answer that was greater than 1 seeing as the region is a subset of the unit square, haha?

I got a very sketchy answer of 0.2529, but I may have done something wrong in my haste.

EDIT: I think I did it wrong...I got 0.266 the second time.

EDIT 2: Final Answer: 0.267 which is actually 4/15 without rounding errors.

 

[keemy]

\( \lim_{x \to 0} \frac{ \ln (1-x) -\sin {x} } {1 - \cos^2{x}} \) oMg

 

[joey]

keemy: -1

 

[qqwref]

A problem with no nasty calculus or algebra:

Spoiler

It's a unit square. B and F are midpoints. Find the area of the shaded region.

Spoiler

Let G be the intersection of BD and CF, and H and I as the other vertices of the shaded region so it is labeled region DGHI.
- Shaded region = ACDE - CGD - EID - AHC - AHE
- Area of shaded region = 1 - 2*(area of CGD) - 2*(area of AHC)
To find the area of those two triangles, let's establish a coordinate system where E = (0,0), D = (1,0), A = (0,1).
Now G is at the intersection of lines y = -2x + 2 and y = x/2 + 1/2. Thus G = (3/5, 4/5), so it's a distance of 2/5 from line CD, and the area of CGD is 1/2 (1)(2/5).
Similarly, H is at the intersection of lines y = x/2 + 1/2 and y = 2x. Thus H = (1/3, 2/3), so it's a distance of 1/3 from line AC, and the area of AHC is 1/2 (1)(1/3).

Finally we get:
- Area of shaded region = 1 - 2*(1/5) - 2*(1/6)
- Area of shaded region = 4/15.

 

[vcuber13]

How exactly did you get an answer that was greater than 1 seeing as the region is a subset of the unit square, haha?

I got a very sketchy answer of 0.2529, but I may have done something wrong in my haste.

EDIT: I think I did it wrong...I got 0.266 the second timev

EDIT 2: Final Answer: 0.267 which is actually 4/15 without rounding errors.

ya um i multiplied by 10 so it is 100 times bigger than i meant, so it should be like .236564

 

[LewisJ]

For something less proof-centered than quad's last problem:

For what values of M, N can an MxN checkerboard be tiled with dominoes? Explain.