What is the Relationship Between X and Y in these Mathematical Sequences?

[anemone]

Find the relationship between $X$ and $Y$ if \(\displaystyle X= 1 + \frac{2}{1!}- \frac{2^3}{3!}+ \frac{2^5}{5!}- \frac{2^7}{7!}+\cdots ...\) and \(\displaystyle Y= 2-\frac{4}{3!}+\frac{6}{5!}-\frac{8}{7!}+\frac{10}{9!}-\cdots ...\)

 

[lfdahl]

My suggested solution:

Spoiler

Comparing:
\[X = 1 + \frac{2^1}{1!}-\frac{2^3}{3!}+\frac{2^5}{5!}-\frac{2^7}{7!}+...\]

- with the Taylor expansion of $sinx$ :

\[sinx =^* x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]

- implies: $X = 1 + sin(2)$

Rewriting the other sum series,$ Y$:

\[Y = 2 - \frac{4}{3!}+\frac{6}{5!}-\frac{8}{7!}+... \\\\ = 2 - \left ( \frac{3}{3!}+\frac{1}{3!} \right )+\left ( \frac{5}{5!}+\frac{1}{5!} \right )-\left ( \frac{7}{7!}+\frac{1}{7!} \right )+... \\\\ = 2 - \left ( \frac{1}{2!}+\frac{1}{3!} \right )+\left ( \frac{1}{4!}+\frac{1}{5!} \right )-\left ( \frac{1}{6!}+\frac{1}{7!} \right )+... \\\\ = \sum_{i=0}^{\infty}\frac{(-1)^i}{2i!} + \sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)!} \\\\ = cos(1)+sin(1)\]

In the last identity (and at (*)) , I have used the following two series (warmest regards to wikipedia ;) ):

\[cosx= \sum_{i=0}^{\infty}\frac{(-1)^ix^{2i}}{2i!} \\\\ sinx = \sum_{j=0}^{\infty}\frac{(-1)^jx^{2j+1}}{(2j+1)!}\]

Thus, $Y = cos(1)+sin(1)$

Therefore, the relation between $X$ and $Y$ is: $X = Y^2$.

 

[Albert1]

My suggested solution:

Spoiler

Comparing: \[X = 1 + \frac{2^1}{1!}-\frac{2^3}{3!}+\frac{2^5}{5!}-\frac{2^7}{7!}+...\]- with the Taylor expansion of $sinx$ :\[sinx =^* x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]- implies: $X = 1 + sin(2)$Rewriting the other sum series,$ Y$:\[Y = 2 - \frac{4}{3!}+\frac{6}{5!}-\frac{8}{7!}+... \\\\ = 2 - \left ( \frac{3}{3!}+\frac{1}{3!} \right )+\left ( \frac{5}{5!}+\frac{1}{5!} \right )-\left ( \frac{7}{7!}+\frac{1}{7!} \right )+... \\\\ = 2 - \left ( \frac{1}{2!}+\frac{1}{3!} \right )+\left ( \frac{1}{4!}+\frac{1}{5!} \right )-\left ( \frac{1}{6!}+\frac{1}{7!} \right )+... \\\\ = \sum_{i=0}^{\infty}\frac{(-1)^i}{2i!} + \sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)!} \\\\ = cos(1)+sin(1)\]In the last identity (and at (*)) , I have used the following two series (warmest regards to wikipedia ;) ):\[cosx= \sum_{i=0}^{\infty}\frac{(-1)^ix^{2i}}{2i!} \\\\ sinx = \sum_{j=0}^{\infty}\frac{(-1)^jx^{2j+1}}{(2j+1)!}\]Thus, $Y = cos(1)+sin(1)$Therefore, the relation between $X$ and $Y$ is: $X = Y^2$.

nice and remarkable!

 

[topsquark]

Y is the letter in the alphabet just after X.

-Dan

 

[anemone]

My suggested solution:

Spoiler

Comparing:
\[X = 1 + \frac{2^1}{1!}-\frac{2^3}{3!}+\frac{2^5}{5!}-\frac{2^7}{7!}+...\]

- with the Taylor expansion of $sinx$ :

\[sinx =^* x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]

- implies: $X = 1 + sin(2)$

Rewriting the other sum series,$ Y$:

\[Y = 2 - \frac{4}{3!}+\frac{6}{5!}-\frac{8}{7!}+... \\\\ = 2 - \left ( \frac{3}{3!}+\frac{1}{3!} \right )+\left ( \frac{5}{5!}+\frac{1}{5!} \right )-\left ( \frac{7}{7!}+\frac{1}{7!} \right )+... \\\\ = 2 - \left ( \frac{1}{2!}+\frac{1}{3!} \right )+\left ( \frac{1}{4!}+\frac{1}{5!} \right )-\left ( \frac{1}{6!}+\frac{1}{7!} \right )+... \\\\ = \sum_{i=0}^{\infty}\frac{(-1)^i}{2i!} + \sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)!} \\\\ = cos(1)+sin(1)\]

In the last identity (and at (*)) , I have used the following two series (warmest regards to wikipedia ;) ):

\[cosx= \sum_{i=0}^{\infty}\frac{(-1)^ix^{2i}}{2i!} \\\\ sinx = \sum_{j=0}^{\infty}\frac{(-1)^jx^{2j+1}}{(2j+1)!}\]

Thus, $Y = cos(1)+sin(1)$

Therefore, the relation between $X$ and $Y$ is: $X = Y^2$.

Very well done, lfdahl, and thanks for participating!

Y is the letter in the alphabet just after X.

-Dan

LOL! Very funny, Dan!