Author Topic: Recreational Problems (SM level) (Read 98419 times) Tweet Share

Mao · « **Reply #135 on:** April 22, 2008, 08:34:33 pm »

Quote from: gfb on April 22, 2008, 07:25:37 pm

Determine whether the given function is a solution of the given initial value problem:

$x(t)=(1+t)^6 (1+t)x'=6x, x(0)=1$
$y(x)=rt(2lnx+x^2-1) xy'=(1+x^2)/y, y(1)=0$

what =\

what is the question actually asking?

is it:
is $(1+t)x'=6x, x(0)=1$ a solution to $x(t)=(1+t)^6$ ??

gfb · « **Reply #136 on:** April 22, 2008, 08:47:04 pm »

It's asking whether x(t)= (1+t)^6 is a solution of (1+t)*x' =6x, x(0)=1

EDIT: Thanks for your interest guys

. midas_touch ( Edward ) has solved the questions.

Ahmad · « **Reply #137 on:** April 22, 2008, 09:08:54 pm »

The title may be a slight misnomer, however, I'd like it if we kept the problems here outside of what you would normally find in a homework problem set, or exam, unless it's a challenge question.

Thanks.

Neobeo · « **Reply #138 on:** May 04, 2008, 04:17:42 pm »

Here's a question I randomly created while attempting to solve another problem:

Find $a\in\mathbb{R}$ such that $y=a^x$ is tangent to $y=x^3$ .

/0 · « **Reply #139 on:** May 04, 2008, 04:56:47 pm »

Quote from: Neobeo on May 04, 2008, 04:17:42 pm

Here's a question I randomly created while attempting to solve another problem:

Find $a\in\mathbb{R}$ such that $y=a^x$ is tangent to $y=x^3$ .

At the tangent point

$a^x = x^3$ (1)

And

$\frac{d}{\,dx} a^x = \frac{d}{\,dx} x^3\Rightarrow \ln{(a)} a^x = 3x^2$ (2)

From the first equation we have

$a = \sqrt[x]{x^3}$ (1')

And substituting into the second:

$\Rightarrow \ln{(\sqrt[x]{x^3})} \times (\sqrt[x]{x^3})^x = 3x^2$

$\ln{(\sqrt[x]{x^3})} \times x^3 = 3x^2$

$\ln{(\sqrt[x]{x^3}) }= \frac{3}{x}$

$\ln{(x^{\frac{3}{x}})} = \frac{3}{x}$

$\frac{3}{x} \ln {x} = \frac{3}{x}$

$\therefore x = e$

$\Rightarrow a^e = e^3$

$\therefore a = \sqrt[e]{e^3}$

Neobeo · « **Reply #140 on:** May 04, 2008, 05:00:23 pm »

Nicely done =).

Neobeo · « **Reply #141 on:** May 04, 2008, 06:29:03 pm »

Find the distance between the graphs of $y=e^x$ and $y=-e^{-x}$ .

dcc · « **Reply #142 on:** May 04, 2008, 07:55:53 pm »

$D = \sqrt{(x_{2} - x_{1})^2 + (-e^{-x_{2}} - e^{x_{1}})^2}$

$D = \sqrt{(x_{2} - x_{1})^2 + (e^{-x_{2}} + e^{x_{1}})^2}$

$\dfrac{\partial D}{\partial x_{1}} = \dfrac{-2(x_{2} - x_{1}) + 2e^{x_{1}}(e^{-x_{2}} + e^{x_{1}})}{2 \sqrt{(x_{2} - x_{1})^2 + (e^{-x_{2}} + e^{x_{1}})^2}}$

$\dfrac{\partial D}{\partial x_{2}} = \dfrac{2(x_{2} - x_{1}) - 2e^{-x_{2}}(e^{-x_{2}} + e^{x_{1}})}{2 \sqrt{(x_{2} - x_{1})^2 + (e^{-x_{2}} + e^{x_{1}})^2}}$

$\dfrac{\partial D}{\partial x_{1}} = 0 = x_{1} - x_{2} + e^{2x_{1}} + e^{x_{1} - x_{2}} \dots \left(1\right)$

$\dfrac{\partial D}{\partial x_{2}} = 0 = x_{2} - x_{1}- e^{-2x_{2}} - e^{x_{1} - x_{2}} = x_{1} - x_{2} + e^{-2x_{2}} + e^{x_{1} - x_{2}} \dots \left(2\right)$

Setting 1 = 2, we get:

$e^{2x_{1}} = e^{-2x_{2}}$

$x_{1} = -x_{2}$

So when D is at a minimum, we can say:

$D = \sqrt{(-2x_{1})^2 + (2e^{x_{1}})^2}$

$D = \sqrt{4x_{1}^2 + 4e^{2x_{1}}}$

$\dfrac{dD}{dx} = \dfrac{8x_{1} + 8e^{2x_{1}}}{2\sqrt{4x_{1}^2 + 4e^{2x_{1}}}}$

Setting this to zero:

$\dfrac{dD}{dx} = 0 = x_{1} + e^{2x_{1}}$

$\implies x_{1} = -e^{2x_{1}}$

By Lambert W:

$x_{1} = -\dfrac{W(2)}{2}$

$x_{2} = \dfrac{W(2)}{2}$

Therefore the minimum distance is:

$D = \sqrt{W(2)^2 + (e^{-\frac{W(2)}{2}} + e^{\frac{-W(2)}{2}})^2}$

$D = \sqrt{W(2)^2 + 4e^{-W(2)}} = \sqrt{W(2)^2 + 2W(2)}$

AppleXY · « **Reply #143 on:** May 04, 2008, 08:01:22 pm »

Wow. I like that.

Neobeo · « **Reply #144 on:** May 07, 2008, 10:38:15 pm »

Neobeo challenges Ahmad to a game. First, Neobeo chooses $n \geq 3$ , and reveals it to Ahmad. Then Neobeo chooses an $n$ -digit number that is divisible by $n$ , but does not reveal this number. Ahmad is then given the task of swapping 2 of these digits, without knowing what Neobeo's $n$ -digit number is. The final number is then revealed and if it is still divisible by $n$ then Ahmad wins, else Neobeo wins. Does either party have a winning strategy? (Can Neobeo choose values such that he is guaranteed a win? Can Neobeo choose values such that Ahmad is not guaranteed a win?)

Example: Neobeo chooses n = 5, then picks the number 12345. Ahmad then asks to swap the first and third digits, producing 32145. It is still divisible by 5 so Ahmad wins.

Edit: Rephrased ambiguous part(s).

bigtick · « **Reply #145 on:** May 08, 2008, 10:55:30 am »

Quote from: Neobeo on May 04, 2008, 06:29:03 pm

Find the shortest distance between the graphs of $y=e^x$ and $y=-e^{-x}$ .

The shortest distance is along the common normal to both graphs. Because of symmetry the common normal passes through the origin.

Let (p,q) lies on $y=e^x$ , .: $q=e^p$ ,

equation of normal at (p,q) is $y-q=-e^{-x}(x-p)$ ,

.: $-q=pe^{-p}$ ,

.: $p=-e^{2p}=-w(2)/2$ , $q^2=\frac{w(2)}{2}$ .

$\mbox{Shortest distance}^2=2^2(p^2+q^2)=w(2)^2+2w(2)$

Neobeo · « **Reply #146 on:** May 11, 2008, 03:09:50 pm »

Well, for the above problem, either Ahmad or Neobeo has a sure-win strategy. Luck doesn't play a factor here. If no one wants to attempt it then I'll put up the solution if someone requests it.

For now though, I've been inspired by polky and the following quote from IRC:

Quote from: IRC, 11th May 2008

[13:43:45] <polky> they split the customers up into red orange and green customers
[13:44:27] <mark_alec> where do you work?
[13:44:32] <polky> safewayyy

So you begin with a safewayyy that is empty except for a single red customer. A red customer has the ability to split into two customers - one orange and one green. Similarly, an orange customer can split into a red and a green customer. Likewise, a green customer can split into a red and an orange customer. Can this safewayyy ever have the same number of red, orange and green customers? (Of course, polky had locked the doors beforehand so none can ever escape this cursed safewayyy.)

Mao · « **Reply #147 on:** May 11, 2008, 04:14:43 pm »

(excuse the big and messy writing)

If cutting was permitted:

1 red -> 1/2 red
-> 1/2 red -> 1/2 green
-> 1/2 orange

dcc · « **Reply #148 on:** May 11, 2008, 04:50:48 pm »

Different Proof:

Assume there is a situation W, where the number of customers in each color are equal.

$\mbox{Let W } = (N, N, N) : N \in \mathbb{Z^+} : (\mbox{Red, Green, Orange}).$

Through the process of unsplitting customers, we can see that the following situation can arise:

$\mbox{W(N - 1, N - 1, N + 1)} \rightarrow \mbox{W(N - 2, N - 2, N + 2)} \rightarrow \mbox{W(N - N, N - N, N + N)} = \mbox{W(0, 0, 2N)}$

Yet this situation is impossible, as when customers are split, the number of customers in each other color type increases. The only way to get to this situation would be for one of these three other situations to exist:

$\mbox{W(-1, 1, 2N - 1) or W(1, -1, 2N - 1) or W(-1, -1, 2N + 1)}$

The number of customers cannot be negative, so $\mbox{(0, 0, 2N)}$ cannot arise, therefore a winning situation cannot be attained either. This contradicts our original assumption that there exists a situation like W, which displays that there is no situation where the number of customers in each color are equal.

Ahmad · « **Reply #149 on:** May 11, 2008, 08:25:00 pm »

(R,G,O) = (1,0,0) = 100 initially. Consider parity, any transformation maps 100 to 011 or vice versa. Therefore 000 or 111 can not be achieved.

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Author Topic: Recreational Problems (SM level) (Read 98419 times) Tweet Share

Mao

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