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The Proof is Trivial! Watch

1. (Original post by Felix Felicis)
x
Very good but it would be perfect if you fixed the typos for clarity!
2. (Original post by und)
Very good but it would be perfect if you fixed the typos for clarity!
Damn, only just noticed how many I made Hope it's all done now
3. Problem 37*/**

Let for denote the sum of the infinite geometric series whose first term is with common ratio .

Evaluate
4. Solution 37

Clearly with hence:

when

and for

Hence it all telescopes and the desired sum is

Really Felix?
5. (Original post by Lord of the Flies)
Really Felix?
6. Time for some polynomial banter.

Problem 38*

Show that there is no polynomial with integer coefficients such that for distinct integers

Problem 39*

Show that if is a polynomial satisfying for all then
7. Solution to Problem 38:

Spoiler:
Show
p(p(p(a))) = p(p(b)) = p(c) = a

p(p(p(b) = p(p(c)) = p(a) = b

p(p(p(c) = p(p(a)) = p(b) = c

Apart from f(x) = x, no such function exists where f(f(f(...f(x)...))) = x (I think)?
8. (Original post by ThatRandomGuy)
Solution to Problem 38:

Spoiler:
Show
p(p(p(a))) = p(p(b)) = p(c) = a

p(p(p(b) = p(p(c)) = p(a) = b

p(p(p(c) = p(p(a)) = p(b) = c

Apart from f(x) = x, no such function exists where f(f(f(...f(x)...))) = x (I think)?
Spoiler:
Show
Those equalities do not imply f(f(f(x))) = x for all x, only for three fixed integers a,b,c - so there is nothing preventing such a function from existing!
9. Problem 40*

Find all functions which satisfy for all real numbers and .
10. (Original post by Noble.)

Problem 27 **/*** (the ** rating very loosely)

Let . Determine

Also, show that your result is consistent with the fact the volume of the unit sphere is
Could you give me a hint of how to go about evaluating this integral, because whichever coordinate system I use, I eventually run into something nasty that I don't know what to do with. It's been driving me mad...

Part 1:

If I try cartesian/cylindrical coordinates, then:
Spoiler:
Show
The region in that we are integrating over is the volume enclosed by the unit sphere , and so

and thus I reach a dead end because of the square root term in the brackets.

If I then try a similar approach with cylindrical coordinates and define the region that we are integrating over as then a very similar dead end occurs in which I have to then evaluate

and again the square root term in the brackets is a problem.

So then I tried spherical coordinates, and:

Spoiler:
Show
The region in that we are integrating over is and so

Let just to make things more manageable and so:

Two iterations of IBP on the integral in square brackets and simplifying yields:

which again is a nightmare of a dead end.

Where am I going wrong?

Part 2:

Spoiler:
Show
Since the region in that we are integrating over is enclosed by the unit sphere, then if we call its volume , then

and comparing with

,

it's obvious that we require and so if I was able to get a result for the first part, I'd substitute this wherever it appears and it should hopefully evaluate to
11. (Original post by Star-girl)
Can you give me a hint of how to go about evaluating this integral, because whichever coordinate system I use, I eventually run into something nasty that I don't know what to do with. It's been driving me mad...

Part 1:

If I try cartesian/cylindrical coordinates, then:
Spoiler:
Show
The region in that we are integrating over is the region enclosed by the unit sphere , and so

and thus I reach a dead end because of the square root term in the brackets.

If I then try a similar approach with cylindrical coordinates and define the region that we are integrating over as then a very similar dead end occurs in which I have to then evaluate

and again the square root term in the brackets is a problem.

So then I tried spherical coordinates, and:

Spoiler:
Show
The region in that we are integrating over is and so

Let just to make things more manageable and so:

Two iterations of IBP on the integral in square brackets and simplifying yields:

which again is a nightmare of a dead end.

Where am I going wrong?

Part 2:

Spoiler:
Show
Since the region in that we are integrating over is enclosed by the unit sphere, then if we call its volume , then

and comparing with

,

it's obvious that we require and so if I was able to get a result for the first part, I'd substitute this wherever it appears and it should evaluate to
Haha, it is a bit of a horrible question!

Hint:

Spoiler:
Show
You can rewrite the integral as

Where

Then, the vector

is of unit length so can be extended to an orthonormal basis with associated coordinates , where:

The unit sphere is still given as as are orthonormal and so:

Which you can now change to spherical polar cooridinates.
12. (Original post by DJMayes)
Problem 36: */**

A particle is projected from the top of a plane inclined at an angle to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection must satisfy:

Is measured from the horizontal or from the plane?
13. (Original post by Felix Felicis)
Is measured from the horizontal or from the plane?
The horizontal. Will edit that back in to make it clear.
14. (Original post by Noble.)
Haha, it is a bit of a horrible question!

Hint:

Spoiler:
Show
You can rewrite the integral as

Where

Then, the vector

is of unit length so can be extended to an orthonormal basis with associated coordinates , where:

The unit sphere is still given as as are orthonormal and so:

Which you can now change to spherical polar cooridinates.
A bit horrible is an understatement...

Oooooh - thanks. Well I could have sat here and scratched my head but I would not have thought of that!

EDIT: I'll write up the full solution another time.
15. (Original post by und)
Problem 40*

Find all functions which satisfy for all real numbers and .
Spoiler:
Show

x=y=0 gives f(0)=0
x=y gives f(x^3)=xf(x^2)
x=-y gives f(x^3)=-f(-x^3)
..I'm sure I only need one or two more to show it can only be f(x)=x (or -x), but I can't think what. Blast!
16. Solution 40

and set

hence where
17. My thoughts on Q39:

Spoiler:
Show

Firstly, notice that if P is a polynomial of odd degree, then P will tend towards negative infinity for either sufficiently large positive or negative x, faster than the other polynomials, so the statement cannot hold true. This means P is a polynomial of even degree with the first term having a positive co-efficient, so that it doesn't tend towards negative infinity, and must have a minimum turning point. At this turning point, P' = 0, P'' is greater than or equal to 0 as it is a minimum. Ignoring the third derivative, it's easy to see this imples P is not less than zero. However, I can't see how to bring the third derivative into this argument.

18. (Original post by DJMayes)
Problem 36: */**

A particle is projected from the top of a plane inclined at an angle to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection from the horizontal must satisfy:

Solution 36
Here you go beautiful

Solution 36

Let denote the horizontal displacement and the projectile is launched with intial velocity v.

, ,

The maximum occurs when

as required

19. Problem 41***

A (countably) infinite class of Cambridge students who believe in the Axiom of Choice are taking their finals for the Mathematical Tripos. At the start of the exam, the examiner places either a white or a black hat on each student at random. There is only one question in the exam: "What colour is your hat?" If only a finite number of students answers incorrectly then everyone becomes a Wrangler (very good), otherwise the Wooden Spoon is given to the whole class (very bad). Everyone can see the hats of everyone else besides their own, and since it's exam conditions they are not allowed to communicate. Students may not remove their hats or try to look at it in any way.

What strategy could the students devise together so to avoid failure and the impending doom of unemployment? Being clever clogs from Cambridge, memorising infinitely large amounts of data is no problem to the students.

Note: A student suggested that each person just guess at random. However this has already been ruled out by the cleverer students who realised that if they were really unlucky, infinite of them could guess wrong.
20. Solution 38

WLOG . Now , and .
Hence and . Consequently, , , which is a contradiction.

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Updated: December 11, 2017
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