Sequences and limit

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Source: Vietnam Mathematical OLympiad 2014

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1077 posts

#1 h Jan 3, 2014, 10:32 AM • 3 Y

Y by NYY, Adventure10, Mango247

Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and
$\begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases}$ for all $n=1,2,3,\ldots$ .
Prove that they are converges and find their limits.

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15142 posts

#2 h Jan 3, 2014, 5:24 PM • 6 Y

Y by pco, Pirkuliyev Rovsen, centslordm, Adventure10, and 2 other users

Assume $x_n = 2\sin \theta_n$ and $y_n = 2\cos \theta_n$ for some $\theta_n \in (0,\pi/2)$ . Then $x_{n+1}^2 = 2(1-\cos \theta_n) = 4\sin^2 \dfrac {\theta_n}{2}$ , so $x_{n+1} = 2\sin \dfrac {\theta_n}{2}$ (the sequences have positive terms), and then $y_{n+1} = \dfrac {x_n} {x_{n+1}} = \dfrac { 2\sin \theta_n} {2\sin \dfrac {\theta_n}{2}} = \dfrac { 2\sin \dfrac {\theta_n}{2}\cos \dfrac {\theta_n}{2}} {\sin \dfrac {\theta_n}{2}} =2\cos \dfrac {\theta_n}{2}$ .

Since $x_{1} = 1 = 2\sin \dfrac {\pi}{6}$ and $y_{1} = \sqrt{3} = 2\cos \dfrac {\pi}{6}$ , the conclusion painlessly follows (the limits are $0$ , respectively $2$ ).

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1748 posts

#3 h Jan 3, 2014, 6:45 PM • 1 Y

Y by Adventure10

why $-2\le {{x}_{n}}\le 2$ ?

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15142 posts

#4 h Jan 3, 2014, 6:51 PM • 3 Y

Y by Nikpour, Adventure10, and 1 other user

Well, we did not know that beforehand, but starting with the given values, it follows that any time $x_1,y_1 > 0$ and $x_1^2+y_1^2 = 4$ , meaning we can take $x_1 = 2\sin \theta$ and $y_1 = 2\cos \theta$ , by induction we will get $x_n = 2\sin \dfrac {\theta} {2^{n-1}}$ and $y_n = 2\cos \dfrac {\theta} {2^{n-1}}$ .

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mathuz

#5 h Feb 4, 2014, 6:46 AM • 2 Y

Y by Adventure10, Mango247

@mavropnevma,
We have the part 'positive sequences'!
So, $y_n\le 2$ and $x_n\le \sqrt{2}$

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8847 posts

#6 h Oct 9, 2021, 2:15 AM • 1 Y

Y by centslordm

The requested limits are $\lim_{n \to \infty} x_n = 0$ and $\lim_{n \to \infty} y_n = 2$ . Let $x_i = 2 \sin a, y_i = 2 \cos a$ -- observe $x_0^2 + y_0^2=4$ . We claim that $x_{i+1} = 2\sin \frac a2, y_{i+1} = 2\cos \frac a2.$ Indeed, verify by double angle that $x_{n+1} = \sqrt{2-y_n} \implies x_{n+1} = \sqrt{2-2 \cos a} = 2\sqrt{\frac{1-\cos a}2} = 2 \sin \frac a2,$ and $x_{n+1} = \frac{x_n}{y_{n+1}} = \frac{2\sin a}{2 \cos \frac a2} = 2\sin \frac a2$ by the double angle formula. But $x_1 = 2\sin \frac{\pi}6$ and $y_1 = 2\cos \frac{\pi}6$ , so as $a$ approaches 0, the two limits are 0 and 2.

This post has been edited 2 times. Last edited by HamstPan38825, Oct 9, 2021, 2:20 PM

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#7 h May 30, 2022, 2:49 PM

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Let $x_n=2a_n$ and $y_n=2b_n$ for all $n \geq 1$ , so $a_1=1/2$ , $b_1=\sqrt{3}/2$ , and we have $a_n=2a_{n+1}b_{n+1}$ and $b_n=1-2a_n^2$ . Recognizing this, as well as the fact that $a_1=\sin(\pi/3)$ and $b_1=\cos(\pi/3)$ , we now claim that $a_n=\sin(2^{1-n}\pi/3)$ and $b_n=\cos(2^{1-n}\pi/3)$ . This holds, as we have $\sin(2^{1-n}\pi/3)=2\sin(2^{-n}\pi/3))\cos(2^{-n}\pi/3))$ and $\cos(2^{1-n}\pi/3)=1-2\sin(2^{-n}\pi/3)^2$ by double angle. Then $\lim_{n \to \infty} x_n=2\lim_{n \to \infty} a_n=2\lim_{n \to \infty}\sin(2^{1-n}\pi/3)=2\sin(0)=0$ , and likewise $\lim_{n \to \infty} y_n=2\cos(0)=2$ . $\blacksquare$

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1352 posts

#8 h Sep 1, 2022, 8:11 PM

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solution

This post has been edited 1 time. Last edited by channing421, Sep 1, 2022, 8:11 PM

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5532 posts

#9 h Jan 30, 2023, 3:14 PM

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We show that $\lim_{n\to \infty} x_n = 0, \lim_{n\to \infty} y_n = 2$ .

Claim: $x_n^2 + y_n^2 = 4$ for all positive integers $n$ .
Proof: We induct on $n$ (base case $n=1$ is obvious). Suppose it was true for everything up to $t$ . We show $x_{t+1}^2 + y_{t+1}^2 = 4$ .

Notice that $x_{t+1}^2= 2 - y_t$ , and $y_{t+1}^2 = \frac{x_t^2}{x_{t+1}^2} = \frac{4 - y_t^2}{2 - y_t} = 2 + y_t$ Hence $x_{t+1}^2 + y_{t+1}^2 = (2-y_t) + (2 + y_t) = 4$ . $\square$

Therefore, let $x_n = 2\sin a_n, y_n = 2 \cos a_n$ , where $a_n\in (0,\pi/2)$ for all positive integers $n$ .

We get $x_{n+1}^2 = 4\sin ^2 a_{n+1} = 2(1-\cos a_n)$ , so $2\sin^2 a_{n+1} = 1 - \cos a_n$ . Since the sequence is positive, we have $a_{n+1} = a_n/2$ by half angle.

As $n\to \infty$ , we have $a_n\to 0$ , so $x_n$ tends to $0$ and $y_n$ tends to $2$ .

This post has been edited 2 times. Last edited by megarnie, Jan 30, 2023, 3:17 PM

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#10 h Mar 20, 2023, 9:47 PM

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Note that $x_1=2\sin 30^\circ$ , $x_2=\frac{\sqrt{3}-1}{\sqrt{2}}=2\sin 15^\circ$ , $y_1=2\cos 30^\circ$ , and $y_2=2\cos 15^\circ$ . So we want to prove that $x_n=2\sin \left(\frac{30}{2^{n-1}}\right)$ , $y_n=2\cos\left(\frac{30}{2^{n-1}}\right)$ .

Suppose that $x_n$ , $y_n$ are as above, and let's show that this pattern holds for $x_{n+1}$ , $y_{n+1}$ .
$x_{n+1}^2=2-y_n=2-2\cos\left(\frac{30}{2^{n-1}}\right)\Longleftrightarrow x_{n+1}=2\sin\left(\frac{30}{2^n}\right).$ Also use the first equation to get $y_{n+1}=\frac{x_n}{x_{n+1}}\Longleftrightarrow y_{n+1}=2\cos\left(\frac{30}{2^n}\right)$ so
$\lim_{n\rightarrow\infty}x_n=0 \text{ and }\lim_{n\rightarrow \infty}y_n=2.$ $\blacksquare$

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#12 h May 25, 2023, 10:07 PM

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Very nice problem!

motivation

solution

This post has been edited 1 time. Last edited by Danielzh, May 25, 2023, 10:08 PM

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eibc

#13 h Dec 30, 2023, 7:34 PM

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I claim that $x_n = 2 \sin(\pi/(3 \cdot 2^n))$ and $y_n = 2 \cos(\pi/(3 \cdot 2^n))$ . We will prove this by induction on $n$ . For $n = 1$ , this is true by definition. Else, assume it holds for $n$ ; we will show it holds for $n + 1$ . Using the second given equation, we have
$x_{n + 1}^2 = 2 - y_n = 2 - 2 \cos \left(\frac{\pi}{3 \cdot 2^n}\right) \implies \left(\frac{x_{n + 1}}{2}\right)^2 = \sqrt{\frac{1 - \cos(\pi/(3 \cdot 2^n))}{2}}.$ By the sine half angle formula, this implies that $x_{n + 1} = 2 \sin (\pi/(3 \cdot 2^{n + 1})).$ For $y_{n + 1}$ , we use the first given equation to get
$y_{n + 1} = \frac{x_n}{x_{n + 1}} = \frac{2 \sin(\pi/(3 \cdot 2^n))}{2 \sin(\pi/(3 \cdot 2^{n + 1}))} = 2 \cos(\pi/(3 \cdot 2^{n + 1})),$ using the sine double angle formula. This completes the induction.

Now, we can see that $\lim_{n \to \infty}x_n = 2 \sin(0) = 0$ and $\lim_{n \to \infty}y_n = 2 \cos(0) = 2$ .

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Jndd

#14 h Sep 2, 2024, 4:54 PM

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We find that $x_1=1=2\cos{60}$ and $x_2=\frac{\sqrt{6}-\sqrt{2}}{2}=2\cos(75)$ , as well as $y_1=\sqrt{3}=2\sin{60}$ and $y_2=\frac{\sqrt{6}+\sqrt{2}}{2}=2\sin{75}$ . Now, we proceed by induction. Suppose that $y_n = 2\sin\theta$ and $x_n=2\cos\theta$ , then $x^2_{n+1}=2-y_n = 2-2\cos(90-\theta) = 2-2(2\cos^2(45-\theta/2)-1)=4\sin^2(45-\theta/2),$ giving $x_{n+1}=2\cos(45+\theta/2)$ . Now, since $y_{n+1}=\frac{x_n}{x_{n+1}} = \frac{\cos\theta}{\cos(45+\theta/2)}$ , we see that $y_{n+1} = 2\sin(45+\theta/2)$ works because $2\sin(45+\theta/2)\cos(45+\theta/2) = \sin(90+\theta)=\cos(-\theta)=\cos(\theta).$ Thus, $(x_n,y_n)=(2\cos\theta, 2\sin\theta)\implies (x_{n+1},y_{n+1}) = (2\cos(45+\theta/2), 2\sin(45+\theta/2)).$ Notice that if $\theta < 90$ , then $45+\theta/2 < 90$ , and also $\theta/2 < 45$ , giving $\theta < 45+\theta/2$ . Hence, the sequence of $\theta$ s is increasing, yet bounded above by $90$ , meaning this sequence converges. Then, we need $\theta = 45 + \theta/2$ , giving $\theta=90$ , so $\lim_{n\to\infty}x_n = 2\cos(90) = 0, \lim_{n\to\infty}y_n = 2\sin(90) = 2.$

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Trasher_Cheeser12321

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Trasher_Cheeser12321

#15 h Dec 20, 2024, 8:40 AM

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Claim: If $x_n = 2\sin(\theta)$ and $y_n = 2\cos(\theta)$ , then we have
$x_{n+1} = 2\sin\left(\frac{\theta}{2}\right) \qquad \text{and} \qquad y_{n+1} = 2\cos\left(\frac{\theta}{2}\right)$ Proof: Plugging in $x_n$ and $y_n$ into the recursive definitions, we're left with the system of equations
$\begin{align*} x_{n+1}y_{n+1} - 2\sin(\theta) &= 0\\ x_{n+1}^2 + 2\cos(\theta) &= 2 \end{align*}$ We can solve from the second equation that $x_{n+1} = \sqrt{2-2\cos(\theta)} = 2\cos\left(\frac{\theta}{2}\right)$ . Substituting this into the first equation and applying double angle formula, we have
$\cos\left(\frac{\theta}{2}\right)y_{n+1} = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$ which simplifies to get us $y_{n+1} = 2\cos\left(\frac{\theta}{2}\right)$ . $\square$

Lastly, since $x_1 = 2\sin\left(\frac{\pi}{6}\right)$ and $y_1 = 2\cos\left(\frac{\pi}{6}\right)$ ,
$\begin{align*} \lim_{n \to \infty} x_n = \lim_{n \to \infty} 2\sin\left(\frac{\frac{\pi}{6}}{2^n}\right) &= \boxed{0} \qquad \text{and}\\ \lim_{n \to \infty} y_n = \lim_{n \to \infty} 2\cos\left(\frac{\frac{\pi}{6}}{2^n}\right) &= \boxed{2} \end{align*}$

This post has been edited 1 time. Last edited by Trasher_Cheeser12321, Dec 20, 2024, 8:42 AM

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eg4334

#16 h Today at 2:17 AM

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when you compute $x_2$ and see the trig the entire problem falls apart. consider $x_n = 2 \sin{\theta}, y_n = 2 \cos{\theta}$ . It's not hard to see by plugging in that $x_{n+1} = 2 \sin{\frac{\theta}{2}}, y_{n+1} = 2 \cos{\frac{\theta}{2}}$ . then the inside of the trig goes to zero and the limit of $x_n$ is $0$ and the limit of $y_n$ is $2$

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