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The Proof is Trivial!

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Reply 2900
Avatar for ctrls
ctrls
9
Original post by Lord of the Flies
∏p(1−1p2)(1+1p(p+1))=∏p(1−1p3)\displaystyle \prod_p \left(1-\frac{1}{p^2}\right) \left(1+\frac{1}{p(p+1)}\right)=\prod_p \left(1-\frac{1}{p^3}\right)


I thought that series look familiar. Seems a bit wrong for me to finish the problem off since you've pointed out the main part, but here goes,

Solution 477

I will assume the Euler product fomula, which states that ζ(s)=∏p11−p−s\displaystyle \zeta(s)= \prod_{p} \frac{1}{1-p^{-s}}, where ζ(s)\displaystyle \zeta(s) denotes the Riemann zeta function. It follows from this formula that 1ζ(s)=(∏p11−p−s)−1=∏p(1−1ps)\displaystyle \frac{1}{\zeta(s)}=\left( \prod_p \frac{1}{1-p^{-s}}\right)^{-1} = \prod_p \left( 1 - \frac{1}{p^s}\right).

Since (1−1p2)(1+1p(p+1))=(1−1p3),\displaystyle \left(1-\frac{1}{p^2}\right) \left(1+\frac{1}{p(p+1)}\right) = \left(1-\frac{1}{p^3}\right), we have ∏p(1+pp+1)=∏p(1−1p3)∏p(1−1p2)=ζ(3)ζ(2)\displaystyle \prod_p \left(1+\frac{p}{p+1}\right) = \frac{\prod_p \left(1-\frac{1}{p^3}\right)}{\prod_p \left(1-\frac{1}{p^2}\right)} = \frac{\zeta(3)}{\zeta(2)} as required.

Just as a note, I don't have any experience with handling infinite series beyond high school level stuff, so I apologize if my manipulations are wrong and/or not fully argued. I just recognized the series and put together a solution.
Original post by ctrls
I thought that series look familiar. Seems a bit wrong for me to finish the problem off since you've pointed out the main part, but here goes,

Solution 477

I will assume the Euler product fomula, which states that ζ(s)=∏p11−p−s\displaystyle \zeta(s)= \prod_{p} \frac{1}{1-p^{-s}}, where ζ(s)\displaystyle \zeta(s) denotes the Riemann zeta function. It follows from this formula that 1ζ(s)=(∏p11−p−s)−1=∏p(1−1ps)\displaystyle \frac{1}{\zeta(s)}=\left( \prod_p \frac{1}{1-p^{-s}}\right)^{-1} = \prod_p \left( 1 - \frac{1}{p^s}\right).

Since (1−1p2)(1+1p(p+1))=(1−1p3),\displaystyle \left(1-\frac{1}{p^2}\right) \left(1+\frac{1}{p(p+1)}\right) = \left(1-\frac{1}{p^3}\right), we have ∏p(1+pp+1)=∏p(1−1p3)∏p(1−1p2)=ζ(3)ζ(2)\displaystyle \prod_p \left(1+\frac{p}{p+1}\right) = \frac{\prod_p \left(1-\frac{1}{p^3}\right)}{\prod_p \left(1-\frac{1}{p^2}\right)} = \frac{\zeta(3)}{\zeta(2)} as required.

Just as a note, I don't have any experience with handling infinite series beyond high school level stuff, so I apologize if my manipulations are wrong and/or not fully argued. I just recognized the series and put together a solution.


The only thing stopping this being totally rigorous is showing that the product you need to find actually converges. Other than that it's dead on.
Original post by arkanm
I've written it up in tex and taken a print screen:



(Note: {x}\{x\} denotes the fractional part of x.)

Can anyone prove this?


I think you must have made an error, there, since that is trivially true.
Original post by james22
I think you must have made an error, there, since that is trivially true.

I thought that initially, but thinking about it it doesn't actually seem that trivially true, especially the inequalities at the end. Even if it is trivially true, the proof seems like it could be interesting.
Original post by arkanm
Please post your proof, I must be experiencing some major brain fart at the moment.



Original post by Jammy Duel
I thought that initially, but thinking about it it doesn't actually seem that trivially true, especially the inequalities at the end. Even if it is trivially true, the proof seems like it could be interesting.


I missread the floor/ceiling functions as absolute values, which would make it trivial.
Original post by james22
I missread the floor/ceiling functions as absolute values, which would make it trivial.

That would have made it beyond trivial.
Original post by arkanm
The problem is equivalent to the Riemann Hypothesis. I have found a truly marvelous demonstration of this equivalence, which this margin is too narrow to contain.


PDF it and send it to me. Thanks. :wink:
Original post by arkanm
I posted it on stackexchange and someone's just posted a very nice solution using continued fractions and stuff. gg folks, gg.

I don't know the consequences of the problem though. Well it certainly solves the trinity admissions quiz problem for good:

"I can divide the number 11 up in a number of ways, eg
11 = 9 + 2
11 = 3.1 + 4.5 + 3.2 + 0.2
but what I want is the way of splitting it up so that the product of all the numbers I split it into is as large as possible. Eg for 11 = 9 + 2 the product concerned is 9 × 2 = 18. Can you find a general way of solving this puzzle so that I could have started with numbers other than 11?"

But it seems to be very natural, and the inequality is sharp. It mysteriously boils down to the fact that e/6<1/2e/6<1/2.


Do you mind linking it?
Not particularly challenging but most get this question wrong.

A bat and a ball cost £1.10, the bat costs £1 more than the ball. How much does the ball cost?
Original post by Excuse Me!
Not particularly challenging but most get this question wrong.

A bat and a ball cost £1.10, the bat costs £1 more than the ball. How much does the ball cost?


Why is that even remotely tricky?

5pence...
Original post by Excuse Me!
Not particularly challenging but most get this question wrong.

A bat and a ball cost £1.10, the bat costs £1 more than the ball. How much does the ball cost?


I saw this on 9gag last night.

I'm ashamed to say I fell for it and said the ball cost 10p.

brb dropping out of my degree in shame.
Original post by In One Ear
Why is that even remotely tricky?

5pence...


Can't remember the exact reason (saw this question a while ago) but its to do with how your brain sees the information. As i said 'most' will get it wrong, not all.
Reply 2912
Avatar for 0x2a
0x2a
2
Problem 478*

(i) Prove that if
∣x−x0∣<min(ϵ2(∣y0∣+1),1) |x - x_0| < min\left(\dfrac{\epsilon}{2(|y_0| + 1)} , 1\right) and ∣y−y0∣<ϵ2(∣x0∣+1)|y - y_0| < \dfrac{\epsilon}{2(|x_0| + 1)} .

then ∣xy−x0y0∣<ϵ|xy - x_0y_0| < \epsilon .

*Note that: min(a,b)=amin(a,b) = a if a<b a < b .

(ii) Prove that if y0≠0y_0 \neq 0 and

∣y−y0∣<min(∣y0∣2,ϵ∣y0∣22)|y - y_0| < min\left(\dfrac{|y_0|}{2},\dfrac{\epsilon|y_0|^2}{2}\right) .

then y≠0 y \neq 0 and

∣1y−1y0∣<ϵ \left|\dfrac{1}{y} - \dfrac{1}{y_0}\right| < \epsilon .

(iii) Replace the question marks with expressions involving ϵ,x0,y0 \epsilon, x_0, y_0 so that the conclusion will be true:

If y0≠0y_0 \neq 0 and

∣y−y0∣< ? |y - y_0| < \ ? and ∣x−x0∣< ?|x - x_0| < \ ?

then y≠0 y \neq 0 and

∣xy−x0y0∣<ϵ\left|\dfrac{x}{y} - \dfrac{x_0}{y_0}\right| < \epsilon.
(edited 9 years ago)
Original post by 0x2a
Problem 478*

(i) Prove that if
∣x−x0∣<min(ϵ2(∣y0∣+1),1) |x - x_0| < min\left(\dfrac{\epsilon}{2(|y_0| + 1)} , 1\right) and ∣y−y0∣<ϵ2(∣x0∣+1)|y - y_0| < \dfrac{\epsilon}{2(|x_0| + 1)} .

then ∣xy−x0y0∣<ϵ|xy - x_0y_0| < \epsilon .

*Note that: min(a,b)=amin(a,b) = a if a<b a < b .

(ii) Prove that if y0≠0y_0 \neq 0 and

∣y−y0∣<min(∣y0∣2,ϵ∣y0∣22)|y - y_0| < min\left(\dfrac{|y_0|}{2},\dfrac{\epsilon|y_0|^2}{2}\right) .

then y≠0 y \neq 0 and

∣1y−1y0∣<ϵ \left|\dfrac{1}{y} - \dfrac{1}{y_0}\right| < \epsilon .

(iii) Replace the question marks with expressions involving ϵ,x0,y0 \epsilon, x_0, y_0 so that the conclusion will be true:

If y0≠0y_0 \neq 0 and

∣y−y0∣< ? |y - y_0| < \ ? and ∣x−x0∣< ?|x - x_0| < \ ?

then y≠0 y \neq 0 and

∣xy−x0y0∣<ϵ\left|\dfrac{x}{y} - \dfrac{x_0}{y_0}\right| < \epsilon.


Been reading spivak? :smile:
Reply 2914
Avatar for 0x2a
0x2a
2
Original post by jjpneed1
Been reading spivak? :smile:

Mhm :tongue:
Prove that a sequence of n integers must contain a sub sequence whose sum is divisible by n?

The proof I know makes use of the pigeonhole principle which isn't taught at A-Level but it very obvious and can be proved very easily.
Original post by alex2100x
Prove that a sequence of n integers must contain a sub sequence whose sum is divisible by n?

The proof I know makes use of the pigeonhole principle which isn't taught at A-Level but it very obvious and can be proved very easily.


We did this in one of our problem classes. :smile:

Spoiler



At the end, when I have like x{p+1} it's meant to be subscript.
(edited 9 years ago)
Original post by rayquaza17
We did this in one of our problem classes. :smile:

Spoiler



At the end, when I have like x{p+1} it's meant to be subscript.


Try learning some TeX - if you just google TeX commands you should get several websites with lists of them. In this case, you just write:

[]x_{p+1}

Where the word "tex" is inserted in each pair of square brackets, to obtain:

xp+1 x_{p+1}
(edited 9 years ago)
[QUOTE="DJMayes;51107335"]Try learning some TeX - if you just google TeX commands you should get several websites with lists of them. In this case, you just write:

xp+1[/tex][br][br]Withoutthespaceinthesecondtag,toobtain:[br][br][tex]xp+1[/tex]x_{p+1} [ /tex][br][br]Without the space in the second tag, to obtain:[br][br][tex] x_{p+1} [/tex]

I know how to do it, I just couldn't be bothered messing about and I thought it still makes enough sense without it. :smile:
Problem 479 **

A die is thrown until an even number appears. What is the expected value of the sum of all the scores?

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