DAILY MATHS CHALLENGE #2

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And the NT problem.

>>8574812
For the calc one:
For each n >= 1, [math]x_{n+1}^3 = x_n^3 + 1 + \frac{1}{3x_n^3} + \frac{1}{27x_n^6}[/math]. Hence, we can write [math]x_n^3 \ge n + \sum_{j=1}^{n-1}\frac{1}{3x_j^3}[/math].
To complete the proof, we need only prove that [math]\sum_{j=1}^{n-1}\frac{1}{x_j^3}[/math] is unbounded. But we have [math]x_{n+1}^3 \le x_n^3 + 3[/math], hence [math]\forall n \ge 1, x_n^3 \le 3n[/math].
Finally, we get [math]\sum_{j=1}^{n-1}\frac{1}{x_j^3} \ge \sum_{j=1}^{n-1} \frac{1}{3j}[/math], which concludes.

Good initiative. I'll gladly ponder both problems. Too few people in this board value problem-solving over memes

>>8574812
different insight (which is overkill but meh)
The sequence is trivially increasing and unbounded (boundedness would yield convergence to 0, which is a contradiction),
hence [math]\frac{1}{x_n} [/math] converges to 0.
Note that [eqn] \begin{align} x_{n+1}^3 &= \left(x_n + \frac{1}{3 x_n^2} \right)^3\\
&=x_n^3\left( 1+\frac{1}{3x_n^3}\right)^3\\
&=x_n^3 \left(1 + \frac{1}{x_n^3}+o\left(\frac{1}{x_n^3}\right) \right)\\
&=x_n^3+1+o(1)
\end{align}[/eqn]
Hence [math] x_{n+1}^3-x_n^3[/math] converges to 1 and Cesaro mean theorem yields [math]x_n^3 = n+o(n)[/math].

Since [math] x_{n+1}^3 = x_n^3 + 1 + \frac{1}{3x_n^3} + \frac{1}{27x_n^6} [/math], [math]x_{n+1}^3 = n+1 + \frac 13 \sum_{k=1}^n \left( \frac 1{x_k^3} + \frac 1{3x_k^6}\right) [/math]
and it suffices to prove that [math] \sum_{k=1}^n \left( \frac 1{x_k^3} + \frac 1{3x_k^6}\right)[/math] is unbounded which is clear since [math]\frac 1{x_n^3} = \frac{1}{n} + o\left( \frac{1}{n} \right) [/math]

>>8574813
Its only true for all even Numbers M. We can write a(k) for a generall M as a(0) for another M'=M^(2k), so we only have to look at the first iteration. Wich only gets integer if M is even

>>8575036
wrong, for example with [math]M=9[/math], [math]a_4 = 2789204756584545 [/math]

>>8575036
Ok, its not completely correct, but the general idea should be right, ill write the Tex when i get home

>>8574813
This is true for M equal to all numbers in the union of the following sets:

S[1] = {2, 4, 6, 8, ...}
S[2] = {3, 7, 11, 15, ...}
S[3] = {5, 13, 21, 29, ...}
S[4] = {9, 25, 41, 57, ...}
...
S[n] = 2S[n-1] - 1
...

Which might be the entire set of natural numbers excluding 1, but I don't know for sure.

I've proved there's an integer term for any M which is not [math] 1 \pmod 4[/math], but I'm stumped dealing with the case [math] 1 \pmod 4[/math] ...

>>8575084
That's where I am too, I'm guessing you have to separate the case M = 1 mod 8 and M = 5 mod 8.
Presumably the M=5 mod 8 case will be easy to prove directly and then we have to look mod 16 for the M=1 mod 8, etc.

>>8574859
I hope you understand you're not working with infinite series.

>>8575084
That's because it works for all integers not [math] \equiv_{4} 1 [/math]. Consider some [math] x \in \mathbb{Z} [/math] such that [math] x \equiv_{4} 1 [/math]. So [math] x = 4k+1,k\in \mathbb{Z} [/math]. This implies [math] a_0 = \frac{8k+3}{2} [/math]. ... I was going somewhere with this, but I can't quite put it into words that other people would be able to follow. The idea is to show that the floor term is always 1 modulo 4, so the numerator will always remain 3 modulo 8, preventing the floor from ever being even, which produces the integer term of the sequence.

>>8575096
samefag, that'll work.
We'll prove by induction that for each [math]k \ge 1[/math], if [math]M \equiv 2^{k-1} + 1 \mod 2^k[/math], then (a_n) has an integer term.
Case k=1: We write [math]M = 2q[/math]. Then, [math]a_1 = \frac{2q+1}{2}[/math] and [math]a_2 = q(2q+1) \in \mathbb N[/math].

Now, assume it's true for some [math]k \ge 1[/math] and let [math]M = 2^{k+1}q+2^k+1[/math]:
We write [math]a_2 = \frac{(2^{k+2}q+2^{k+1}+3)(2^{k+1}q+2^k+1)}{2} = \frac{2N+1}{2}[/math], where the numerator satisfies [math]2N+1 \equiv 3(2^k +1) \equiv 2^k+3 \mod 2^{k+1}[/math], and thus [math]N \equiv 2^{k-1}+1 \mod 2^k[/math]. Hence, by induction hypothesis, the sequence [math](a_{k+1})_{k \ge 1}[/math] contains an integer term.
But each integer M greater than 1 satisfies [math]M \equiv 2^{k-1}+1 \mod 2^k[/math] for some [math]k \ge 1[/math].
Indeed, if [math]k = \min\{k \ge 1, M \not \equiv 1 \mod 2^k\}[/math], then since [math]M \equiv 1 \mod 2^{k-1}[/math] by definition, we can write [math] M = 2^{k-1}q + 1[/math].
Then, since [math]M \not \equiv 1 \mod 2^k[/math], we have [math]2 \not | q[/math]. Hence we can write [math] q = 2 q'+1[/math] and finally [math]M = 2^kq' + 2^{k-1} + 1[/math].
Hence, for each integer M > 1, the sequence (a_n) with a_1 = (2M+1)/2 has an integer term.
It doesn't for M=1 since it is constant with value 3/2.

>>8575104
I do, but I'm not sure I understand where you're going with this

>>8574812

would anyone suggest a basic book on proofs to me?

I would love to learn how to solve stuff like this.

Here's a calc problem since both previous problems have been solved:
Let [math]f: [a,b] \to \mathbb C[/math] a piecewise continuously differentiable function. Assume [math]f(a) = f(b) = 0[/math] and [math]|f'(x)| \le M[/math] wherever it makes sense.
Prove that [math]\displaystyle \left|\int_a^b f(t)dt\right| \le M\frac{(b-a)^2}{4}[/math].
What's the equality case ?

>>8575449

does it involve the taylor expansion?

>>8575449

i think the equality case would be a triangle-shaped piecewise linear function.

>>8575577
Yes, also I wanted to clarify what I mean by piecewise continuously differentiable: by that I mean a continuous function that is an antiderivative of a piecewise continuous function

>>8575636
that's right ! got a proof ?

Provide a bijection of (0, 1) into [0, 1].

It's possible.

>>8575655
Let
[math] N=\{ \frac{1-3^{-k}}{2}, 3^{-k}+\frac{1-3^{-k}}{2} \mid k\in \mathbb{N}\cup \{0\} \} [/math]
[math] M=[0,1]-N [/math]

and so we get the bijection
[math] \varphi: [0,1] \to (0,1)[/math] defined by
[math] \varphi(x)= \begin{cases} (x+1)/3 & x \in N \\ x & x \in M \end{cases} [/math]

>>8575655
Send 0 to 1/2, 1 to 1/3, 1/n to 1/(n+2) for each n >= 2 and leave everyone else fixed. That gives you a bijection from [0,1] to (0,1).
An interesting follow-up question might be to prove that you can't find a continuous bijection, one way or the other

>>8575663
How did you come up with such a convoluted formulas? Did you know the answer already?

>>8575669
>How did you come up with such a convoluted formulas?
an algorithmic proof of Cantor–Schröder–Bernstein

>Did you know the answer already?
yes i posted the same bijection a week or so ago

its not really that convoluted, you just contract the endpoints along an infinite sequence similarly to the way >>8575664 did and leave the rest fixed

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>>8575664
What's a continuous bijection?
>>8575673
> an algorithmic proof of Cantor–Schröder–Bernstein
So basically picrel?

>>8575686
Well a bijection [math]f: [0,1] \to (0,1)[/math] (or the other way around) that is also a continuous function in the usual sense

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>>8575686
yes basically

>>8575655
1 to 0.9 to 0.99 to 0.99 etc
0 to 0.1 to 0.11 to 0.11 etc
Everything else the same

Easy.

Let [math] K [/math] be an algebraic system with two binary operations (one written additively, the other multiplicatively), satisfying:

(1) [math] K [/math] is an abelian group under addition,
(2) [math] K - \{0\} [/math] is a group under multiplication, and
(3) [math] x(y+z) = xy + xz [/math] for all [math] x, y, z \in K [/math] .

Suppose that for some [math] n [/math] , [math] 0 = 1 + 1 + ... + 1 [/math] ( [math] n [/math] times). Prove that, for all [math] x \in K [/math] , [math] (-1)x = -x [/math] .

>>8575664
Suppose such an function f exists. If f mapped [0,1] to (0,1) this would imply (0,1) is compact. Similarly if f mapped (0,1) to [0,1] it would imply that the inverse image of [0,1] under f is closed, which it clearly is not.

>>8575664
In fact you only need that f is a surjection, injection is unnescessary.

>>8575836
>Similarly if f mapped (0,1) to [0,1] it would imply that the inverse image of [0,1] under f is closed, which it clearly is not.
It's closed in (0,1), there's no contradiction. You need more.

>>8575841
Injection is unnecessary for the [0,1] -> (0,1) case, but it's necessary for the (0,1) -> [0,1] case (you can find continuous surjections from (0,1) to [0,1])

>>8575871
Ok so if the bijection from (0,1) to [0,1] was continuous then it would imply the inverse is continuous which is not.

>>8575941
yup

>>8575779
needs fresher material, not old kacysnski papers posted so recently

>>8575995

You get a cookie for identifying the provenance. :^)

I notice, however, that you also have not proffered a solution.

>>8576086
its more fun doing questions youve never seen before

>>8575449
Assuming the function is continuously differentiable, and [math]0\leq a\leq b[/math],
integration by parts yields [eqn]\left|\int_a^b f(t)dt\right| = \left|\int_a^b tf'(t)dt\right|\leq M\frac{b^2-a^2}2 [/eqn] which is not as sharp as what's required (because of my additional assumptions on f)

>>8576667
That's not the reason it wasn't as sharp as what's required, you just went a wee bit too fast ;)
Also, don't worry, you can integrate by parts under the weaker hypothesis I put, but there's a way of finding this identity that does not require you to integrate by parts

>>8576680
yes, my estimate can be sharpened with a variational method. Since [math]\int_a^b f'(t)dt =0[/math], for any [math]\alpha[/math] we have
[eqn]\left|\int_a^b f(t)dt\right| = \left|\int_a^b (t-\alpha)f'(t)dt\right|\leq M \int_a^b |t-\alpha| dt [/eqn]
Some calculus or educated guess shows that [math]\alpha=\frac{a+b}2[/math] minimizes [math]\int_a^b |t-\alpha| dt [/math] and plugging in [math]\alpha=\frac{a+b}2[/math] yields the result

>>8576710
Very nice, quite different from my solution but I like yours better because it puts the problem in a larger context.
Mine was a collection of tricks:
Using Fubini and the fundamental theorem of calculus, we rewrite:
[math]\displaystyle \int_a^b f(t)dt = \int_a^b \int_a^t f'(u) du dt = \int_a^b (b-u)f'(u)[/math]
[math]\displaystyle \int_a^b f(t)dt = - \int_a^b \int_t^b f'(u) du = \int_a^b (a-u)f'(u)du[/math]
And hence:
[math]\displaystyle \int_a^b f(t) dt = \int_a^b \left(\frac{a+b}{2}-u\right) f'(u)du [/math]
And then [math]\displaystyle \left|\int_a^b f(t) dt \right| \le \int_a^b |f'(u)| \left| \frac{a+b}{2}-u\right| du \le M \frac{(b-a)^2}{4}[/math]
The equality case in the second inequality is when |f'| is equal to M, and the equality case in the first inequality is exactly when f' and a+b/2-u have opposite signs, which tells us that a function that satisfies the equality case must be symmetric relative to a+b/2 and triangle shaped

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Alright, my turn to suggest a problem

This one is from the latest issue of the College Mathematics Journal. You can submit your solution until February iirc

>>8575753
1 is gonna become 1 after continuum worth of being functioned. So no. Retarded piece of shit.

>>8576928
>>8576928
You've misused Fubini's Theorem in lines 1 and 2, and your integral in the fourth line converges to 0. The inequality is still valid however.

>>8578240
The Fourth line actually converges to ab, which may or may not be greater than or equal to the upper bound presented In the problem.

>>8578258
I meant -ab. Without the constraint that both a and b are greater than or equal to 0, the inequality doesn't necessarily hold.

>>8578240
>You've misused Fubini's Theorem in lines 1 and 2
How ? [math]\displaystyle \int_a^b \left(\int_a^t f'(u)du\right)dt = \int_a^b \left(\int_u^b f'(u)dt\right) du[/math]

>The Fourth line actually converges to ab
what ?

>>8576981
Hmm I can't make it till the end...

Fubini's Theorem doesn't permit you to integrate dt over the domain of du. dt strictly corresponds to the domain a to b and du from a to t.

> The Fourth line ...
Do the integral.

Nevermind about the fourth line; I'm retarded.

>>8575779
hint for a compsci person whose never taken algebra?

what additional structure would you normally need to prove (-1)*x=-x without that strange identity 0 = 1+....+1?

>>8579072
It's not "the domain of dt" (or I'm not understanding you right). Let's see it with an extra step:
[eqn]\int_a^b \left(\int_a^t f'(u)du\right) dt = \int_a^b \left(\int_a^b 1_{u \le t}f'(u) du \right) dt \underset{Fubini}{=} \int_a^b \left(\int_a^b 1_{u \le t}f'(u) dt \right) du = \int_a^b \left(\int_u^b f'(u) dt \right) du[/eqn]

>>8579171
You normally use left-distributivity, ie. (x+y)z = xz + yz. Here all you have is right-distributivity

>>8579171
What you normally use is that:
>Assotiativity of addition
>Existence of a neutral element of addition
>Existence of additive inverses of all elements
>Existence of a neutral element of multiplication
>Right distributivity
For all x in K we have
x + (-1)*x
= 1*x + (-1)*x
= (1 + (-1))*x
= 0*x
= 0*x + 0
= 0*x + (0*x + (-(0*x)))
= (0*x + 0*x) + (-(0*x))
= (0+0)*x + (-(0*x))
= 0*x + (-(0*x))
= 0.
Analogously (-1)*x + x = 0.
Therefore (-1)*x = -x.

>>8579189

Pardon me, but you have confused left-distributivity with right-distributivity. Kaczynski's problem only gives you an instance of left-distributivity, and not right-distributivity as you just stated.

https://en.wikipedia.org/wiki/Distributive_property#Definition

http://mathworld.wolfram.com/Distributive.html

in each case, it is the /thing outside the parentheses/ which can informally be said to "distribute" across the two-term expression contained inside. That thing's position is where these two names come from, not the position of the two-term thing inside. After all, the thing inside the parentheses has not more than one term to "distribute" to, in the primitive case of the definitions!

>>8579200
i had done it this way:

(1+-1)x = x + -x
x + (-1)x = x + -x
(-1)x = -x

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>>8579187

>>8579228
autism

>>8579189

i guess the obvious thing that follows from 0=1+...+1 is that any element, combined n times with itself using + is zero, so you have

-x+....+-x = (-1)x + .... + (-1)x

but i can't figure how to make the extra terms go away.

>>8575779

this is tough. any way that you use 0 = 1+....+1 you end up with a mess of extra terms that are hard to deal with.

how long is the proof?