The Proof is Trivial!

Original post by bananarama2

Directly below?

Well yes? What are you wanting us to account for, the rotation of the earth whilst it's being dropped? :lol:

Reply 1041

Original post by Jkn

Well yes?

Well no

Reply 1042

Original post by bananarama2

Well no

Well yes! You have phrased the problem as a physics problem... not a mathematics problem or even a Newtonian Mechanics problem. Unless the effect is measurable and consistent under the parameters you have set out, then we must assume it will land directly below :tongue:

...wait so what you have in mind isn't the rotation of the earth... ? :tongue:

Reply 1043

Original post by Jkn

Well yes! You have phrased the problem as a physics problem... not a mathematics problem or even a Newtonian Mechanics problem. Unless the effect is measurable and consistent under the parameters you have set out, then we must assume it will land directly below :tongue:

...wait so what you have in mind isn't the rotation of the earth... ? :tongue:

You assume nothing. It most certainly doesn't land below and of course it's a Newtonian mechanics problem.

Well that's for you do decide :wink:

Reply 1044

jack.hadamard

4

Original post by Jkn

Edit: Where's this problem from btw? It was rather neat :smile:

Well, I saw it on this webpage. Otherwise, it is USAMO 1994.

Reply 1045

Mladenov

1

Original post by Lord of the Flies

Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.

Original post by ben-smith

Very nice Mlad! I'm too slow again :tongue:

Thanks.

Original post by Jkn

And I don't see what's going on in 159. Where does the first line come from? And how does it link to the second and third? The differentiation thing doesn't seem to make sense (though I'm sure it does), could you clarify it for me :tongue:

My solution looks so different to yours, it's in you form of two symmetrical series expression involving all the square numbers (and no factorials!) :lol:

(I checked it though and it does work!)

The first line is just integration by parts. Then, we represent

\sin^{n}x

as a sum of sines or cosines, depending on the value of

n \pmod 2

.

Problem 163**

Let

f \in C^{1}

. Find

\displaystyle \lim_{n\to \infty} n\left(\int_{0}^{1}f(x)dx - \frac{1}{n}\sum_{i=1}^{n} f(\frac{i}{n}) \right)

.

Problem 164**

Evaluate

\displaystyle I = \int_{0}^{\infty} \frac{\arctan x}{x^{\alpha}}dx

, where

\alpha \in \left(1,2 \right)

.

Problem 165***

Evaluate

\displaystyle \int_{0}^{2} \frac{x^{4}}{(x^{2}+1)(x(2-x)^{3})^{\frac{1}{4}}}dx

.

Problem 166***

Evaluate

\displaystyle \int_{0}^{\infty} \frac{x\ln x}{(1+x^{2})(1+x^{3})^{2}}dx

.

Problem 167*** (one that is interesting, and not trivial)

Evaluate

\displaystyle \sum_{n} \frac{1}{n^{2}}

, over the set of all biquadrate free positive integers.
If you are interested, find

\displaystyle \sum_{n} \frac{(-1)^{\frac{n-1}{2}}}{n}

, over the set of all odd cube free positive integers.

Reply 1046

Original post by jack.hadamard

Well, I saw it on this webpage. Otherwise, it is USAMO 1994.

Awesome cheers! Question 4 looks fun! Btw, are there any UK university maths competitions? :tongue:

Or anything between Oxford and Cambridge colleges and stuff like that? :tongue:

Original post by bananarama2

You assume nothing. It most certainly doesn't land below and of course it's a Newtonian mechanics problem.

Well that's for you do decide :wink:

Reply 1047

Original post by Jkn

The best problems are the simplest :smile:

Reply 1048

jack.hadamard

4

Original post by Mladenov

Problem 167*** (one that is interesting, and not trivial)

I like this. Below is a hint; a question that I had posted in the inappropriate place at the inappropriate time.

Spoiler

EDIT: Nevermind, read it the right way.

(edited 10 years ago)

Reply 1049

jack.hadamard

4

Original post by Jkn

Btw, are there any UK university maths competitions? :tongue:

I don't know of any maths competitions here that can be compared to Putnam.
However, the IMC is an international competition for university students run by UCL.

Reply 1050

Original post by jack.hadamard

I don't know of any maths competitions here that can be compared to Putnam.
However, the IMC is an international competition for university students run by UCL.

Going to Bulgaria this year I see :lol:

Reply 1051

Original post by bananarama2

The best problems are the simplest :smile:

Lets assume you drop it off the second platform of height "h" which is given to be 115m

We take the period of earth's rotation "t" to be 23 hours, 56 minutes and 4 seconds and the radius of the earth to be "r".

Assuming the immediate curvature of the each in negligible, we can assume that the tangential velocity is equal to the horizontal velocity.

For the earth,

V_e=k\frac{2\pi r}{t}

,

For the ball,

V_b=k\frac{2\pi (r+h)}{t}

.

Let v denote the speed of the ball with respect to the ground in a direction tangential to the curvature of the earth.

\Rightarrow v=k\frac{2\pi h}{t} \approx 0.00839k m/s

.

The latitude of Paris is approximately 48.87 N. Hence

k=\cos(48.87)

.

Let the distance the ball lands (in a westerly direction, i.e. under the Eiffel Tower) be denoted by d.

We have that

d=vt

and that

115=\frac{1}{2} g t^2

\Rightarrow d=\cos(48.87) \times 0.00839 \sqrt{\frac{330}{9.81}} \approx 3.20 cm

, to 3 significant figures.

\square

Spoiler

(edited 10 years ago)

Reply 1052

SParm

8

Original post by bananarama2

Problem 162 *

If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)

What do you want us to do? If we assume nothing we have to take into account that we're in a rotating frame of reference and that certainly is not "*". Not for any A-Level mechanics course that I've seen.

Reply 1053

ukdragon37

14

Original post by Jkn

Solution 162

Lets assume you drop it off the second platform of height "h" which is given to be 115m

We take the period of earth's rotation "t" to be 23 hours, 56 minutes and 4 seconds and the radius of the earth to be "r".

Assuming the immediate curvature of the each in negligible, we can assume that the tangential velocity is equal to the horizontal velocity.

For the earth,

V_e=k\frac{2\pi r}{t}

,

For the ball,

V_b=k\frac{2\pi (r+h)}{t}

.

Let v denote the speed of the ball with respect to the ground in a direction tangential to the curvature of the earth.

\Rightarrow v=k\frac{2\pi h}{t} \approx 0.00839k m/s

.

The latitude of Paris is approximately 48.87 N. Hence

k=\cos(48.87)

.

Let the distance the ball lands (in a westerly direction, i.e. under the Eiffel Tower) be denoted by d.

We have that

d=vt

and that

115=\frac{1}{2} g t^2

\Rightarrow d=\cos(48.87) \times 0.00839 \sqrt{\frac{330}{9.81}} \approx 3.20 cm

, to 3 significant figures.

\square

Spoiler

But the base of the Eiffel tower is much wider than the middle or at the top. If you dropped it vertically off the second platform, and it only moved 3.20 cm, how did it reach the bottom without hitting the side or the first platform? :pierre:

Reply 1054

Original post by SParm

What do you want us to do? If we assume nothing we have to take into account that we're in a rotating frame of reference and that certainly is not "*". Not for any A-Level mechanics course that I've seen.

For this instance the rotating reference frame whilst playing a part isn't anything beyond a-level. There is a way of doing it that involves the Coriolis and centrifugal force, but there is a way which doesn't use it.

Reply 1055

Original post by ukdragon37

But the base of the Eiffel tower is much wider than the middle or at the top. If you dropped it vertically off the second platform, and it only moved 3.20 cm, how did it reach the bottom without hitting the side or the first platform? :pierre:

Since when have the compscis become such realists?

Reply 1056