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.

5 years, 3 months ago

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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{align*} \sin 2\theta &= 2 \sin \theta \cos \theta = \frac{336}{625}, \\ \cos 2\theta &= 1 - 2\sin^2 \theta = \frac{527}{625}. \end{align*} Hence \begin{align*} 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{align*} But we also have \begin{align*} 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{align*} 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{align*} k_0 + \frac{24}{25} k_1 &= \frac{527}{625}, \\ \frac{7}{25} k_1 &= \frac{336}{625}, \end{align*} 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.

- 5 years, 3 months ago

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.

- 5 years, 3 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.

Staff - 5 years, 3 months ago

Thanks

- 5 years, 3 months ago