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The Proof is Trivial!

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Reply 160
Original post by Felix Felicis
x

Very good but it would be perfect if you fixed the typos for clarity!
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 :colondollar: Hope it's all done now
Problem 37*/**

Let SrS_{r} for r=1,2,3,...,100r = 1, 2, 3, ..., 100 denote the sum of the infinite geometric series whose first term is r1r!\dfrac{r-1}{r!} with common ratio 1r\dfrac{1}{r}.

Evaluate 1002100!+r=1100(r23r+1)Sr\dfrac{100^{2}}{100!} + \displaystyle\sum_{r=1}^{100} |(r^{2} - 3r + 1) S_{r} |
(edited 11 years ago)
Solution 37

Clearly Sr=1(r1)!S_r= \dfrac{1}{(r-1)!} with S1=0S_1=0 hence:

(r23r+1)Sr=r1(r2)!r(r1)!=r1(r2)!r(r1)!\big| (r^2-3r+1)S_r\big|=\bigg| \dfrac{r-1}{(r-2)!}-\dfrac{r}{(r-1)!}\bigg|=\dfrac{r-1}{(r-2)!}-\dfrac{r}{(r-1)!} when r>2r>2

and =1\cdots =1 for r=2r=2

Hence it all telescopes and the desired sum is 33

Really Felix? :tongue:
(edited 11 years ago)
Time for some polynomial banter.

Problem 38*

Show that there is no polynomial pp with integer coefficients such that p(a)=b,  p(b)=c,  p(c)=ap(a)=b,\;p(b)=c,\;p(c)=a for distinct integers a,b,ca,b,c

Problem 39*

Show that if pp is a polynomial satisfying p(x)+p(x)p(x)+p(x)p(x)+p'''(x)\geq p'(x)+p''(x) for all xx then p(x)0p(x)\geq 0
(edited 11 years ago)
Solution to Problem 38:

Spoiler

(edited 11 years ago)
Original post by ThatRandomGuy
Solution to Problem 38:

Spoiler



Spoiler

Reply 168
Problem 40*


Find all functions f:RRf:\mathbb{R}\to \mathbb{R} which satisfy f(x3)+f(y3)=(x+y)(f(x2)+f(y2)f(xy))f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy)) for all real numbers xx and yy.
Original post by Noble.
Here's an additional problem:

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

Let a,b,cRa,b,c \in \mathbb{R}. Determine

I=x2+y2+z21cos(ax+by+cz) dxdydzI = \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(ax+by+cz) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z

Also, show that your result is consistent with the fact the volume of the unit sphere is 4π/34 \pi/3


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... :tongue:

Part 1:

If I try cartesian/cylindrical coordinates, then:

Spoiler



So then I tried spherical coordinates, and:

Spoiler



Where am I going wrong? :confused:

Part 2:

Spoiler

(edited 11 years ago)
Reply 170
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... :tongue:

Part 1:

If I try cartesian/cylindrical coordinates, then:

Spoiler



So then I tried spherical coordinates, and:

Spoiler



Where am I going wrong? :confused:

Part 2:

Spoiler



Haha, it is a bit of a horrible question!

Hint:

Spoiler

Original post by DJMayes
Problem 36: */**

A particle is projected from the top of a plane inclined at an angle ϕ \phi 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 θ \theta must satisfy:

θ=π4ϕ2 \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}

Is θ\theta measured from the horizontal or from the plane?
Original post by Felix Felicis
Is θ\theta measured from the horizontal or from the plane?


The horizontal. Will edit that back in to make it clear.
Original post by Noble.
Haha, it is a bit of a horrible question!

Hint:

Spoiler



A bit horrible is an understatement... :tongue:

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

EDIT: I'll write up the full solution another time.
(edited 11 years ago)
Reply 174
Original post by und
Problem 40*


Find all functions f:RRf:\mathbb{R}\to \mathbb{R} which satisfy f(x3)+f(y3)=(x+y)(f(x2)+f(y2)f(xy))f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy)) for all real numbers xx and yy.


Spoiler

Solution 40

y=0f(x3)=xf(x2)  ()y=0\Rightarrow f(x^3)=xf(x^2)\;(\star) and set xz(x)=f(x).xz (x)=f(x).

()x3z(x3)=x3z(x2)z(x3)=z(x2)z(x)=C(\star)\Rightarrow x^3z(x^3)=x^3z(x^2)\Rightarrow z(x^3)=z(x^2)\Rightarrow z (x)=\mathcal{C} hence f(x)=Cxf(x)=\mathcal{C}x where C=f(1)\mathcal{C}=f(1)
My thoughts on Q39:

Spoiler

Original post by DJMayes
Problem 36: */**

A particle is projected from the top of a plane inclined at an angle ϕ \phi 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 θ \theta from the horizontal must satisfy:

θ=π4ϕ2 \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}

Solution 36
Here you go beautiful :wink:

36 Solution

(edited 11 years ago)
:colone:

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.
(edited 11 years ago)
Reply 179
Solution 38

WLOG a<b<ca<b<c. Now ba(modca)b \equiv a \pmod {c-a}, cba(modba)c \equiv b \equiv a \pmod {b-a} and bca(modcb)b \equiv c \equiv a \pmod {c-b}.
Hence ba=cab-a=c-a and cb=bac-b=b-a. Consequently, ca=c+b2ac-a=c+b-2a, a=b=ca=b=c, which is a contradiction.

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