Does the problem "Matt's Recurrance" include 0 in the Natural Numbers?

I'm not sure if it would change the problem, but one way would make solving it different for me.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

It does not, I couldn't solve it starting with n+2=2.

Try it without including it. It should've said it in the description, though. It took a lot of time for me to see that.

Mustafa Özçiçek - 8 years, 2 months ago

Thanks

Maddy B - 8 years, 2 months ago

Note that your 'conclusion' is false.

You merely tried to find values which satisfy 1 of the many equations, and did not find a set of values which satisfied ALL of the equations.

Read the solution, which is similar to what Hero posted above.

Calvin Lin Staff - 8 years, 1 month ago

Now that the question is closed, I can respond.

If one solves the recurrence relation $a_{n+2} = k_1 a_{n+1} + k_0 a_n$ for suitable constants $k_0, k_1$ such that it holds for all positive integers $n$ , then it also holds for all real numbers $n$ . Therefore, it makes no difference whether one considers the case $n = 0$ in the solution; either way, the same constants will be obtained.

To see why this is the case, we note that if $\sin \theta = \frac{7}{25}$ , with $0 < \theta < \frac{\pi}{2}$ , then $\cos \theta = \frac{24}{25}$ . Then $\begin{aligned} \sin 2\theta &= 2 \sin \theta \cos \theta = \frac{336}{625}, \\ \cos 2\theta &= 1 - 2\sin^2 \theta = \frac{527}{625}. \end{aligned}$ Hence $\begin{aligned} a_{n+2} &= \sin n \theta \cos 2\theta + \cos n \theta \sin 2\theta \\ &= \frac{527}{625} \sin n\theta + \frac{336}{625} \cos n\theta. \end{aligned}$ But we also have $\begin{aligned} a_{n+2} &= k_1 a_{n+1} + k_0 a_n \\ &= k_1 (\sin n \theta \cos \theta + \cos n \theta \sin \theta) + k_0 \sin n \theta \\ &= \left( k_0 + \frac{24k_1}{25} \right) \sin n\theta + \frac{7k_1}{25} \cos n\theta. \end{aligned}$ So if this recurrence is to be satisfied for positive integers $n$ , then it is immediately obvious that it is also satisfied for arbitrary real $n$ , since the equality of the coefficients gives identical equality. In particular, we must have $\begin{aligned} k_0 + \frac{24}{25} k_1 &= \frac{527}{625}, \\ \frac{7}{25} k_1 &= \frac{336}{625}, \end{aligned}$ the solution of which is $k_0 = -1$ , $k_1 = \frac{48}{25}$ . Therefore, $k_0 + k_1 = \frac{23}{25}$ . However, we should note that if we were to approach the question by substituting in $n = 0$ , the resulting equation is underdetermined for $k_0$ . That doesn't mean the recurrence is false for this value; it simply means that $k_0$ is not uniquely determined in that specific case.

hero p. - 8 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$