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jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Integration Bee Kaizo
Calcul8er   43
N 39 minutes ago by awzhang10
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
43 replies
Calcul8er
Mar 2, 2025
awzhang10
39 minutes ago
Double factorial identity
Snoop76   0
4 hours ago
Source: own
Show that the following identity holds:$$\sum_{k=0}^n (2k+3)!!{n\choose k}=2(n+1)\sum_{k=0}^n (2k+1)!!{n\choose k}+\sum_{k=0}^n (2k-1)!!{n\choose k}$$
0 replies
Snoop76
4 hours ago
0 replies
AB similar to BA
Andyqian7   0
Today at 3:33 PM
Source: CMC Final 2017
Let $A$, $B$ be $n\times n$ matrices such that $\text{rank}(ABA)=\text{rank}(B)$. Prove that $AB$ is similar to $BA$.
0 replies
Andyqian7
Today at 3:33 PM
0 replies
Concavity of a function
pii-oner   1
N Today at 1:50 PM by alexheinis
Hi everyone,

I am studying the concavity of the function

\[
f(x) = \sqrt{1 - x^a}, \quad a \geq 0
\]
on the interval \( x \in [0,1] \).

I computed the second derivative and found that for \( a \geq 1 \), the function appears to be concave. However, I am uncertain about the behavior at the endpoints.

Does anyone have insights on confirming concavity rigorously for \( a \geq 1 \) and understanding the behavior at the endpoints? Any help would be greatly appreciated!

Thanks!
1 reply
pii-oner
Today at 9:00 AM
alexheinis
Today at 1:50 PM
No more topics!
Putnam 2017 B3
goveganddomath   34
N Today at 4:16 AM by OronSH
Source: Putnam
Suppose that $$f(x) = \sum_{i=0}^\infty c_ix^i$$is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational.
34 replies
goveganddomath
Dec 3, 2017
OronSH
Today at 4:16 AM
Putnam 2017 B3
G H J
G H BBookmark kLocked kLocked NReply
Source: Putnam
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goveganddomath
11679 posts
#1 • 4 Y
Y by Adventure10, Rounak_iitr, Mango247, Knight2E4
Suppose that $$f(x) = \sum_{i=0}^\infty c_ix^i$$is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational.
This post has been edited 2 times. Last edited by goveganddomath, Dec 3, 2017, 10:31 PM
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goveganddomath
11679 posts
#2 • 2 Y
Y by Adventure10, Mango247
You can use the denominators to show that there are infinitely many $c_i$ equal to $1$. From there, that shows that if the sum converges with denominators of powers of $3$ to a quantity with a non-common denominator of $2$, it will be irrational for an initial denominator of $2$...that's a rough sketch at least.
This post has been edited 1 time. Last edited by goveganddomath, Dec 3, 2017, 10:31 PM
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62861
3564 posts
#3 • 6 Y
Y by Ravi12346, Binomial-theorem, shashank30122000, Supercali, Adventure10, Mango247
We will show that if $f(1/2)$ is rational, then $f(2/3)$ is a rational with odd denominator.

If
\[f(1/2) = c_0.c_1c_2c_3 \dots_{(2)}\]is rational, then $\{c_i\}$ is eventually periodic: there exist positive integers $N, d$ such that $c_{n + d} = c_n$ whenever $n > N$. Then
\begin{align*}
f(2/3) & = \sum_{i = 0}^{\infty} c_i (2/3)^i\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} c_{N + i} \left((2/3)^{N + i} + (2/3)^{N + d + i} + \cdots \right)\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} \frac{c_{N + i}(2/3)^{N + i}}{1 - (2/3)^d}\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} \frac{c_{N + i}2^{N + i} 3^d}{3^{N + i}(3^d - 2^d)},
\end{align*}a sum of rationals with odd denominators; consequently $f(2/3)$ is a rational with odd denominator. (We may rearrange the terms in the sum because it contains only nonnegative terms and converges; hence converges absolutely.)
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amoghanakru1029
5 posts
#4 • 1 Y
Y by Adventure10
You think if I justified my steps up to the point where i got a $3^d - 2^d$ in the denominator, I was probably on the right track?

I first checked whether the binary decimal was terminating (it wasn't, because of a lemma I derived based on denominators of finite powers of 2/3) and then repeating, but I'm not sure if they'll take off because I didn't explicitly call my first case-check a lemma (but I did reference it).
This post has been edited 1 time. Last edited by amoghanakru1029, Dec 3, 2017, 10:43 PM
Reason: d
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Anzoteh
124 posts
#5 • 2 Y
Y by Adventure10, Mango247
Just wondering, don't you have to show that the sequence $c_i$ has to be periodic for $f(1/2)$ to be rational?
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goveganddomath
11679 posts
#6 • 2 Y
Y by Adventure10, Mango247
$f(1/2)$ is irrational.
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amoghanakru1029
5 posts
#7 • 2 Y
Y by Adventure10, Mango247
Anzoteh wrote:
Just wondering, don't you have to show that the sequence $c_i$ has to be periodic for $f(1/2)$ to be rational?

Yea, that's what CantonMath used to show that it's in fact irrational. You can also check terminating which is a far easier case (especially if you checked repeating first).
This post has been edited 1 time. Last edited by amoghanakru1029, Dec 3, 2017, 10:44 PM
Reason: d
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gaga654
8 posts
#8 • 2 Y
Y by Adventure10, Mango247
Terminating is just a special (period=1) case of periodic
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amoghanakru1029
5 posts
#9 • 2 Y
Y by Adventure10, Mango247
It's probably true that setting f(x) = y for this power series uniquely determines the coefficients $c_i$. Does anyone have results on how setting $f(x) = y$ for rational x, y determines the rationality/irrationality of other f(x)=y?
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jmerry
12096 posts
#10 • 2 Y
Y by Adventure10, Mango247
No, amoghanakru1029, it doesn't determine the coefficients uniquely, at least not for any $x>\frac12$. The contrapositive/base-2 expansion method is the way to go. My proof is essentially identical to CantonMathGuy's.
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Anzoteh
124 posts
#11 • 2 Y
Y by Adventure10, Mango247
amoghanakru1029 wrote:
Anzoteh wrote:
Just wondering, don't you have to show that the sequence $c_i$ has to be periodic for $f(1/2)$ to be rational?

Yea, that's what CantonMath used to show that it's in fact irrational. You can also check terminating which is a far easier case (especially if you checked repeating first).

Yeah I mean can you just quote the fact that "if $f(1/2)$ is rational then the sequence is eventually periodic?" I had most of the proof of this in the exam but I skipped the final computation (basically some Euler's function involved...ew...)
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TheStrangeCharm
290 posts
#12 • 2 Y
Y by Adventure10, busy-beaver
^Proving that a number is rational iff its binary expansion is eventually periodic is not that difficult. If the expansion is eventually periodic, it is pretty clear from geometric series computations that the number is rational. Conversely, let $0 < \alpha < 1$ be rational number. Multiplying and dividing by powers of $2$ does not change whether the sequence is eventually periodic or not, so assume $\alpha = \frac{a}{b}$ with $a$ and $b$ positive integers and gcd($a$, $b$, 2) = $1$. By Fermat's Little Theorem there exists a positive integer $N$ with $2^{N} - 1$ divisible by $b$. Then $(2^{N} - 1)\frac{a}{b}$ is an integer, so the binary expansion of $\alpha$ is periodic with period $N$.
This post has been edited 3 times. Last edited by TheStrangeCharm, Dec 4, 2017, 2:55 AM
Reason: typo3
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amoghanakru1029
5 posts
#13 • 2 Y
Y by Adventure10, Mango247
jmerry wrote:
No, amoghanakru1029, it doesn't determine the coefficients uniquely, at least not for any $x>\frac12$. The contrapositive/base-2 expansion method is the way to go. My proof is essentially identical to CantonMathGuy's.

CantonMathGuy showed that if f(1/2) is rational, then f(2/3) is a rational with odd denominator.

Is my approach valid? I assumed $c_i$ formed a terminating or eventually repeating sequence, and showed that the equation $f(2/3) = 3/2$ was impossible (due to denominator arguments). Then I said by contradiction, f(1/2) is irrational
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jmerry
12096 posts
#14 • 1 Y
Y by Adventure10
amoghanakru1029 wrote:
I assumed $c_i$ formed a terminating or eventually repeating sequence, and showed that the equation $f(2/3) = 3/2$ was impossible (due to denominator arguments
It works. I just prefer the contrapositive formulation over thinking in terms of a contradiction; you're doing essentially the same things, but it might be easier to understand.
After all, where does that assumption come from? It comes from taking $f(\tfrac12)$ rational.
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ernacktob
3 posts
#15 • 2 Y
Y by Adventure10, Mango247
After obtaining the expression as in CantonMathGuy's post, I used the 2-adic valuation to show that $v_2(f(2/3)) \geq 0$ since all fractions have denominators with no powers of $2$ while $v_2(1/2) = -1$. Would that be considered correct? Just in case, I derived the basic theory of valuations like $v_2(x + y) \geq \min(v_2(x), v_2(y))$. But I guess this might actually be an overkill.
This post has been edited 2 times. Last edited by ernacktob, Dec 6, 2017, 8:16 AM
Reason: Added comment.
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silouan
3952 posts
#16 • 1 Y
Y by Adventure10
You can see problem 1 here of SEEMOUS 2007:
https://olimpiadamate.files.wordpress.com/2014/05/2007_seemous.pdf
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djduan
68 posts
#17 • 2 Y
Y by Adventure10, Mango247
CantonMathGuy wrote:
We will show that if $f(1/2)$ is rational, then $f(2/3)$ is a rational with odd denominator.

If
\[f(1/2) = c_0.c_1c_2c_3 \dots_{(2)}\]is rational, then $\{c_i\}$ is eventually periodic: there exist positive integers $N, d$ such that $c_{n + d} = c_n$ whenever $n > N$. Then
\begin{align*}
f(2/3) & = \sum_{i = 0}^{\infty} c_i (2/3)^i\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} c_{N + i} \left((2/3)^{N + i} + (2/3)^{N + d + i} + \cdots \right)\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} \frac{c_{N + i}(2/3)^{N + i}}{1 - (2/3)^d}\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} \frac{c_{N + i}2^{N + i} 3^d}{3^{N + i}(3^d - 2^d)},
\end{align*}a sum of rationals with odd denominators; consequently $f(2/3)$ is a rational with odd denominator. (We may rearrange the terms in the sum because it contains only nonnegative terms and converges; hence converges absolutely.)

Could someone explain to me how it's possible for $f(\frac{2}{3}) = \frac{3}{2}$ in any case? It seems to me that $f(\frac{2}{3})$ is always a sum of rationals with odd denominator anyways.
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jmerry
12096 posts
#18 • 3 Y
Y by djduan, Adventure10, Mango247
Easily. Build it by the greedy algorithm; $c_0=1$, $c_1=0$, $c_2=1$, $c_3=c_4=c_5=c_6=c_7=0$, and so on; $c_n$ is $1$ if the remainder is greater than $\left(\frac23\right)^n$, and zero otherwise.
Going fifty terms out, I get $101000001001001010000000001000000100001000000100100$. (Any farther, and I'm afraid of floating-point errors creeping in)

A sum of (infinitely many) rationals with odd denominator doesn't have to be a rational with odd denominator, because it doesn't have to be rational.
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Mathlete2017
3233 posts
#19 • 2 Y
Y by Adventure10, Mango247
silouan wrote:
You can see problem 1 here of SEEMOUS 2007:
https://olimpiadamate.files.wordpress.com/2014/05/2007_seemous.pdf
The link doesn't seem to work. I know this post is almost a year old, but if you see this, could you fix it?
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Farbe_Bears
19 posts
#20 • 1 Y
Y by Mango247
CantonMathGuy wrote:
We will show that if $f(1/2)$ is rational, then $f(2/3)$ is a rational with odd denominator.

If
\[f(1/2) = c_0.c_1c_2c_3 \dots_{(2)}\]is rational, then $\{c_i\}$ is eventually periodic: there exist positive integers $N, d$ such that $c_{n + d} = c_n$ whenever $n > N$. Then
\begin{align*}
f(2/3) & = \sum_{i = 0}^{\infty} c_i (2/3)^i\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} c_{N + i} \left((2/3)^{N + i} + (2/3)^{N + d + i} + \cdots \right)\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} \frac{c_{N + i}(2/3)^{N + i}}{1 - (2/3)^d}\\
& = \sum_{i = 0}^{N} c_i (2/3)^i + \sum_{i = 1}^{d} \frac{c_{N + i}2^{N + i} 3^d}{3^{N + i}(3^d - 2^d)},
\end{align*}a sum of rationals with odd denominators; consequently $f(2/3)$ is a rational with odd denominator. (We may rearrange the terms in the sum because it contains only nonnegative terms and converges; hence converges absolutely.)

Actually I have a question... do u clear it?
c_i's are the coefficients of the given series. When u put x=1/2 then u have
f(1/2)= \sum_{i = 0}^{\infinty} c_i (1/2)^i
Which is essentially = c_0+(c_1)/2+(c_2)/2²+(c_3)/2³+...
Is it equal to = c_0.c_1c_2c_3 \dots_{(2)}\

??????
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kevinmathz
4680 posts
#21
Y by
Solved with help from pinetree1.

Notice that for all $n$, we have that $v_3\left( \sum_{i=0}^{n} x^ic_i\right) < 0$ because the denominator is always a power of $2$. Now, this means that there are an infinite amount of $c_i$ such that $c_i=1$. Now, we assume the contrapositive: $f(0.5)$ is rational. Writing in binary decimal, this means that there $c_i$ repeats, meaning after a number $n$ then there exists some $k$ such that for all $i\ge n$, $c_i=c_{i+k}$. This means that, letting $S=\sum_{i=0}^{n} x^ic_i$ gets that our sum is $$S+\sum_{j=1}^k \left(\frac{2^{n+j}}{3^{n+j}} c_{n+j}\sum_{l=0}^{\infty} \frac{2^{kl}}{3^{kl}}\right) = S + \sum_{j=1}^k \frac{2^{n+j}}{3^{n+j}} c_{n+j} \cdot \frac{1}{1-\frac{2^k}{3^k}}$$and since that expression's denominators are all off and the sum is finite, then $v_2$ of that expression is nonnegative. However, $v_2(1.5) = -1$ which reaches a contradiction.
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HamstPan38825
8846 posts
#23 • 2 Y
Y by jhu08, Math4Life7
A relatively uninteresting problem?

Assume for the sake of contradiction that $f(1/2)$ is rational. Then, interpret the value of the series as the value of a binary decimal. This decimal represents a rational number if and only if it is eventually periodic, by definition. Hence, the series must be eventually periodic.

Suppose that $\{c_n\}$ is periodic for all $c \geq k$, in particular with $c_k = 1$. Furthermore, let the period of repetition be $r$. Then, our sequence can be split into two parts: the repeating part and the nonrepeating part. Note that by definition, the sum of the repeating part is $$\frac{3^r}{3^r-2^r} \cdot (\text{sum of period of repetition}).$$The $\nu_2$ of the inner expression is at least $k \geq 0$ by definition, so the $\nu_2$ of the entire expression must also be $k$.

Now, consider the nonrepeating part. Obviously the $\nu_2$ of the nonrepeating part is nonnegative; thus, the sum of the series must also have a nonnegative $\nu_2$, contradiction. Thus, $f(1/2)$ is necessarily irrational.
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john0512
4170 posts
#24
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We will show that the sequence $c_i$ cannot be eventually periodic. FTSOC assume otherwise. Then, let $x$ and $p$ be such that for $i\geq x$, we have $$c_{i}=c_{i+p}.$$We then have $$c_0+(2/3)c_1\cdots +(2/3)^{x-1}c_{x-1}+\frac{(2/3)^x c_x + (2/3)^{x+1}c_{x+1}\cdots (2/3)^{x+p-1}c_{x+p-1}}{1-(2/3)^p}=\frac{3}{2}.$$Let's rewrite this: $$c_0+(2/3)c_1\cdots +(2/3)^{x-1}c_{x-1}+\frac{3^p((2/3)^x c_x + (2/3)^{x+1}c_{x+1}\cdots (2/3)^{x+p-1}c_{x+p-1})}{3^p-2^p}=\frac{3}{2}.$$Since $3^p-2^p$ is odd, and all the other denominators on the left side are 3, there is no way to get a even number in the denominator for the LHS, contradiction. Therefore, the sequence cannot be eventually periodic, so $f(1/2)$ is irrational.
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peace09
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This post has been edited 1 time. Last edited by peace09, Sep 8, 2023, 3:12 AM
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joshualiu315
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#27
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If $f(1/2)$ is rational, we have

\[f(1/2)=c_0.c_1c_2c_3\dots_{2}\]
must be periodic. In other words, after a certain point $k$, we have $c_n=c_{n+p}$. Then,

\begin{align*}
f(2/3) &= \sum_{n = 0}^{\infty} c_n(2/3)^n \\
&=\sum_{n=0}^{k} c_n(2/3)^n + \sum_{n=1}^p c_{k+n}((2/3)^{k+n}+(2/3)^{k+p+n}+\dots) \\
& = \sum_{n = 0}^{k} c_n (2/3)^n + \sum_{n = 1}^{p} \frac{c_{k + n}2^{k + n} 3^p}{3^{k + n}(3^p - 2^p)} 
\end{align*}
which clearly has an odd denominator, contradiction. $\square$
This post has been edited 1 time. Last edited by joshualiu315, Jan 5, 2024, 1:21 AM
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Leo.Euler
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#28
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Assume otherwise. Then $c_i$ is periodic. Thus if $k$ is the period of $c_i$, \[ \frac{3}{2} \cdot \left( \frac{3^k-2^k}{3^k} \right) \]is equal to a finite decimal in base $2/3$. But this is impossible since the denominator in reduced form is even, as desired.
This post has been edited 1 time. Last edited by Leo.Euler, Nov 30, 2023, 5:33 PM
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Shreyasharma
666 posts
#29
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Assume for sake of contradiction that both are rational. Now note that if $f\left(\frac{1}{2}\right)$ it has either a terminating or eventually repeating binary expansion (!!). This is in fact the main difficulty of the problem.
\newline

Assume it is terminating, so for all $m > k$ we have $c_m = 0$. However now as $\nu_3\left(\frac{3}{2}\right) = 1$ and $\nu_3 \left(\left(\frac{2}{3} \right)^k\right) = -k$. Now as,
\begin{align*}
    \nu_p(x+y) = \min(\nu_p(x), \nu_p(y))
\end{align*}as long as $\nu_p(x) \neq \nu_p(y)$ so clearly this case fails.
\newline

Now consider the case where $\{c_n\}$ is eventually repeating. Assume it has an initial string of $b+1$ characters, and then repeats with period $m$. Then we need,
\begin{align*}
   \frac{3}{2} = \frac{q}{3^{b}} + \frac{r}{3^{b+m}} \left(1 + \left(\frac{2}{3}\right)^m + \left(\frac{2}{3} \right)^{2m} + \dots \right)
\end{align*}for some integer constants $q$ and $r$ not divisible by $3$. From sum of geometric series we find,
\begin{align*}
    \frac{3}{2} &= \frac{q}{3^b} + \frac{r}{3^{b+m}}\left(\frac{1}{1 - \left(\frac{2}{3} \right)^m} \right)\\
    &= \frac{q}{3^b} + \frac{r}{3^{b}}\left(\frac{1}{3^m -2^m}\right)\\
    &= \frac{1}{3^b} \left(q + \frac{r}{3^m - 2^m} \right)
\end{align*}Then we must have,
\begin{align*}
    \frac{3^{b+1}}{2} = q + \frac{r}{3^m - 2^m}
\end{align*}However now examining the $\nu_2$ we have a contradiction as the right hand side can never have an even denominator.
This post has been edited 2 times. Last edited by Shreyasharma, Dec 22, 2023, 6:03 AM
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Martin.s
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#30
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Suppose \(f\left(\frac{1}{2}\right)\) is explicit. In this case, in its binary expansion (resulting from the given order), we would have a periodicity of coefficients from a point onwards. So there are polynomials with coefficients \(P, Q\) in \(\{0, 1\}\) such that

\[
f(x) = P(x) + \left(1 + x^N + x^{2N} + \ldots\right)Q(x) = P(x) + \frac{Q(x)}{1 - x^N}
\]
For \(x = \frac{2}{3}\), the left member is equal (given) to \(\frac{3}{2}\). But the right, making fractions homonymous separately in \(P, Q\) and adding, is of the form

\[
\frac{A}{3^M} + \frac{B}{(3^N - 2^N)3^K}
\]
with integer \(A, B\). Multiplying now by the least common multiple \(E\) of the two denominators on the right, which is an odd number, follows \(\frac{3E}{2} =\) an integer. Out of place, and so on.
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chakrabortyahan
371 posts
#31
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Cute problem!! $\blacksquare\smiley$oops
Attachments:
This post has been edited 1 time. Last edited by chakrabortyahan, Apr 29, 2024, 6:47 AM
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de-Kirschbaum
184 posts
#32
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Suppose that $f\left ( \frac{1}{2} \right )$ is rational, then its decimals must eventually be periodic in all bases. Considering it in base $2$, for some starting point $d$, some period $k$, some $i \in [0, k-1]$, $c_{d+i}=c_{d+i+kt}$ for all nonnegative integers $t$. Thus, we know that $c_n$ must be eventually periodic, which transforms $f(x)$ into a finite sum.

Consider now $f \left (\frac{2}{3} \right )$. For all $d+i+kt$ we have $c_{d+i} \left ( \frac{2}{3} \right )^{d+i} + \ldots = c_{d+i}\left ( \frac{3^{k-d-i}2^{d+i}}{3^k-2^k} \right )$. That means

$$f \left (\frac{2}{3} \right )=\sum_{i=0}^{d-1}c_i \left (\frac{2}{3} \right )^i + \sum_{i=0}^{k-1} c_{d+i}\left ( \frac{3^{k-d-i}2^{d+i}}{3^k-2^k} \right ) = \frac{3}{2}$$
However, note that the $\nu_2$ of each term in the sequence are $0, 1, 2, \ldots, d-1, d, d+1, \ldots, d+k-1 \geq 0$ at max (if they even contribute any $\nu_2$ to the sum at all). That means the $\nu_2$ of the sequence can never drop below $0$ by the strong triangle inequality. Thus, $\nu_2 \left (f\left (\frac{2}{3} \right )\right ) > \nu_2\left (\frac{3}{2} \right )=-1$, a contradiction.
This post has been edited 1 time. Last edited by de-Kirschbaum, Aug 26, 2024, 6:42 PM
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numbertheory97
39 posts
#33
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Nice problem!

Solution. Suppose to the contrary that $f(1/2)$ is rational. Since $\sum_{n \geq 0} c_n/2^n$ is just the binary representation of $f(1/2)$, it's well-known that the sequence $c_0, c_1, \dots$ is eventually periodic. Suppose the sequence begins to repeat at $c_k$ with period $r$ (it's possible that $k = 0$). Then we have \[f\left(\frac23\right) = \sum_{n = 0}^{k - 1} c_n\left(\frac23\right)^n + \frac{3^r}{3^r - 2^r} \sum_{n = k}^{k + r - 1} c_n\left(\frac23\right)^n,\]so since $\nu_2(3^r/(3^r - 2^r))$ is clearly $0$, we have \[\nu_2(f(2/3)) \geq \min\left(\nu_2\left(\sum_{n = 0}^{k - 1} c_n\left(\frac23\right)^n\right), \nu_2\left(\sum_{n = k}^{k + r - 1} c_n\left(\frac23\right)^n\right)\right) \geq 0,\]a contradiction. $\square$
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bjump
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#34
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Suppose for the sake of contradiction that $f(1/2) \in \mathbb Q$. This is equivalent to the binary decmal $c_0. c_1 c_2 \cdots$ being rational. Therefore this means that $\{ c_i \}$ must be eventually periodic.

Now suppose that $c_i$ has period length and a new cycle starts at $c_{k+1}$. By the geometric series formula we require
$$\sum_{i=0}^k \left(\frac{2}{3}\right)^i c_i + \frac{\sum_{i=k+1}^{k+m} (\tfrac{2}{3})^i c_i}{1-(\tfrac{2}{3})^m}= \frac{3}{2} $$$$\sum_{i=0}^k \left(\frac{2}{3} \right)^i c_i + 3^m \cdot\frac{\sum_{i=k+1}^{k+m} (\tfrac{2}{3})^i c_i}{3^m-2^m}= \frac{3}{2} $$Observe that the LHS is the sum of finitely many fractions with odd denominators. Therefore the LHS can be written as a fraction with an odd denominator. However the RHS has an even denominator. Contradiction.
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Jndd
1417 posts
#35
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FTSOC, suppose $f\left(\frac12\right)$ is rational. This implies the $\{c_n\}$ must be eventually periodic. It can't have a finite number of $1$s clearly, since $f\left(\frac{2}{3}\right)$ would have a denominator that's a power of $3$, so the repeating part must be nonzero. Let $\sum_{n=0}^i c_nx^n$ be the beginning part that doesn't repeat. Then, the part that does repeat can be written as $(c_ix^i+\cdots c_{k+i-1}x^{k+i-1})(x^0+x^k+x^{2k}+\cdots ) = (c_ix^i+\cdots c_{k+i-1}x^{k+i-1})\cdot \frac{1}{1-x^k} = \left(c_i\left(\frac{2}{3}\right)^i+\cdots c_{k+i-1}\left(\frac{2}{3}\right)^{k+i-1}\right)\cdot\frac{3^k}{3^k-2^k}$. Now, it's clear that the denominator of $f\left(\frac{2}{3}\right)$ is odd, so it can't possibly simplify to $\frac{3}{2}$.
This post has been edited 1 time. Last edited by Jndd, Jan 25, 2025, 4:46 AM
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Aiden-1089
277 posts
#36
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By considering the base $2$ representation of $f \left( \frac{1}{2} \right)$ it is clear that $f \left( \frac{1}{2} \right)$ is rational iff the sequence ${c_n}$ is eventually periodic.
Now assume the sequence is eventually periodic. Let $N,T$ be the smallest integers such that $c_k=c_{k+T}$ for all $k \geq N$.
$f \left( \frac{2}{3} \right) = \frac{a}{3^{N-1}} + \left( \frac{2}{3} \right) ^N \cdot \frac{b}{3^{T-1}} \cdot \frac{1}{1-(\frac{2}{3})^T} = \frac{1}{3^{N-1}} \left( a+\frac{2^N b}{3^T-2^T} \right)$ where $0 \leq a \leq 3^N-2^N, 0 \leq b \leq 3^T-2^T$.
$f \left( \frac{2}{3} \right) = \frac{3}{2} \implies 3^N= 2 \left( a+\frac{2^N b}{3^T-2^T} \right)$. But $3^T-2^T$ is always odd, contradiction.
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OronSH
1718 posts
#37
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If $f(1/2)$ is rational then by binary $c_i$ is eventually periodic, so $f(2/3)$ is the sum of finitely many powers of $2/3$, and finitely more powers of $2/3$ times $\frac1{1-(2/3)^k}=\frac{3^k}{3^k-2^k}$ for some $k$. However none of these terms have negative $\nu_2$ so they cannot sum to $3/2$.
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