Waste less time on Facebook — follow Brilliant.
×

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.

Note by Maddy B
4 years, 7 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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

Sort by:

Top Newest

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.

Hero P. - 4 years, 7 months ago

Log in to reply

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 - 4 years, 7 months ago

Log in to reply

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 - 4 years, 7 months ago

Log in to reply

Thanks

Maddy B - 4 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...